Density — Paper 4 · IGCSE Physics 0625

5 (a) Compare the arrangement and motion of the molecules in ice and in liquid water. ice … … water … … [2] (b) An ice-hockey rink has an area of 1800 m2. The ice has a thickness of 0.025 m. The density of ice is 920 kg / m3. (i) Calculate the mass of ice on the rink. mass = … [2] (ii) The ice is at 0 °C. To form the ice, water at 0 °C was poured onto the floor of the rink and then frozen. The specific latent heat of fusion of ice is 3.3 × 105 J / kg. Calculate the energy removed from the water to form the ice at 0 °C. energy = … [2] [Total: 6]

6 marks

Mark scheme: 5(a) Molecular arrangement: Ice: in lattice / regular / arranged / orderly / fixed in place Water: random / irregular / not arranged / not orderly Molecular movement: Ice: vibrate Water: move (around) or slide over each other B2 5(b)(i) d = m / V in any form OR (m =) Vd OR 1800 × 0.025 × 920 = 41 000 kg C1 A1 5(b)(ii) (H =) mL OR 41 400 × 3.3 × 105 1.4 x 1010 J OR 1.4 × 107 kJ OR 1,4 × 104 MJ C1 A1 Total 6

This question in 0625/41 Oct/Nov 2016

5 (a) State a comparison of (i) the arrangement of the molecules in ice and in liquid water, ice … liquid water … [1] (ii) the movement of the molecules of ice and liquid water. ice … liquid water … [1] (b) The mass of the ice on an ice-hockey rink is 51 000 kg. (i) The density of ice is 920 kg / m3. Calculate the volume of ice on the rink. volume = … [2] (ii) To form the ice, water at 0 °C was poured onto the floor of the rink and then frozen. The specific latent heat of fusion of ice is 3.3 × 105 J / kg. Calculate the quantity of energy removed from the water to form this ice at 0 °C. energy = … [2] (c) The temperature of the ice is monitored by a thermometer made up of wires of two different metals connected to a remote voltmeter. The junction of the two wires is embedded in the ice. State the name of this type of thermometer. … [1] [Total: 7]

7 marks

Mark scheme: 5(a)(i) in ice, molecules in fixed positions AND in water, positions change B1 5(a)(ii) in ice, molecules vibrate AND in water, molecules move around (and vibrate) B1 5(b)(i) m / ρ OR ρ = m / V in any form (V = 51 000/920 =) 55 m3 C1 A1 5(b)(ii) mL OR Q = mL in any form (Q = 51 000 × 3.3 × 105) = 1.7 × 1010 J C1 A1 5(c) thermocouple B1

This question in 0625/42 Oct/Nov 2016

5 (a) (i) The temperature of a quantity of air is increased. Describe the effect of this increase in temperature on the air molecules. … … [1] (ii) The temperature of a block of aluminium is decreased. Describe the effect of this decrease in temperature on the density of the aluminium. Explain your answer in terms of the separation of the atoms. effect … explanation … … [2] (b) A road is covered with a layer of snow. The temperature of the snow is 0 °C. The specific latent heat of fusion of snow is 3.3 × 105 J / kg. The snow forms a layer of uniform thickness on the road surface. (i) Calculate the power needed to melt 0.12 kg of the snow in 220 s. power = … [4] (ii) The snow is melted by solar radiation. In 220 s, exactly 0.12 kg of the snow melts to produce water at 0 °C. The power of the solar radiation incident on each 0.12 kg of snow is greater than the power calculated in (b)(i). Suggest one reason why this is so. … … [1] [Total: 8]

8 marks

Mark scheme: 5(a)(i) speed increases or kinetic energy increases B1 5(a)(ii) effect: density / it increases explanation: (separation) decreases and less (kinetic) energy / vibrate less / smaller volume B1 B1 5(b)(i) (Q = )ml or 0.12 × 3.3 × 105 3.96 / 4.0 × 104 (P = )Q / t or ml / t or 3.96 × 104 / 220 ign. WD / t 180 W C1 C1 C1 A1 5(b)(ii) thermal energy lost to something specific (air, road, surroundings) or thermal radiation reflected B1 Total 8

This question in 0625/43 Oct/Nov 2016

4 Fig. 4.1 shows a Galilean thermometer. This thermometer is used to measure the approximate temperature of the surrounding air. bulb E, label: 28 °C glass cylinder bulb D, label: 26 °C bulb C, label: 24 °C bulb B, label: 22 °C water bulb A, label: 20 °C Fig. 4.1 The glass cylinder contains water. When the temperature of the water changes, so does its density. Each bulb has a label printed with a temperature, as shown in Fig. 4.1. The bulbs have different densities. At 21 °C, only bulb A is at the bottom of the cylinder. (a) Explain, in terms of density, why bulb A is at the bottom of the cylinder and the other bulbs are floating. … … … … [2] (b) The temperature of the surrounding air increases to a temperature above 23 °C. (i) Suggest one reason why there is a delay before the temperature of the water increases to 23 °C. … [1] (ii) Explain why, after this delay, bulb B sinks. Assume the bulbs do not expand. … … … … [3] (c) Bulbs A, B and C are now at the bottom of the cylinder. Bulbs D and E are floating. State the possible temperature range of the water in the cylinder. … [1]

7 marks

Mark scheme: 4(a) Density of bulb A greater than the density of the water (and sinks) B1 Density of other bulbs less than the density of water (and float) B1 4(b)(i) Glass is a poor conductor of heat OR glass conducts heat at a slow rate OR water has a high (specific) heat capacity B1 4(b)(ii) The water expands OR separation of water molecules increases B1 The water becomes less dense B1 Bulb B now has a greater density than the water (and sinks) OR Weight of bulb B more than buoyancy forces / upthrust B1 4(c) 24 oC – 26 oC B1 Total: 7

This question in 0625/42 Feb/March 2017

3 Fig. 3.1 shows remote sensing equipment on the surface of a distant planet. Fig. 3.1 (a) The mass of the equipment is 350 kg. The acceleration of free fall on the surface of this planet is 7.5 m / s2. (i) State what is meant by the term weight. … … [1] (ii) Calculate the weight of the equipment on the planet. weight = … [2] (b) The equipment releases a balloon from a point that is a small distance above the surface of the planet. The atmosphere at the surface of this planet has a density of 0.35 kg / m3. The inflated balloon has a mass of 80 g and a volume of 0.30 m3. Make an appropriate calculation and then predict and explain the direction of any motion of the balloon. Show your working. prediction … explanation … … [4] [Total: 7]

7 marks

Mark scheme: 3(a)(i) (Weight is) force/pull of gravity (acting on an object) B1 3(a)(ii) Mass × acceleration due to gravity OR mg OR 350 × 7.5 C1 2600 N A1 3(b) (ρ =) m / V in any form C1 0.27 (kg / m3) OR 270 (g / m3) A1 Balloon moves/floats up B1 (Floats when) density of balloon less than density of atmosphere OR (sinks when) density of balloon greater than atmosphere B1 OR (ρ =) m / V in any form (C1) 110 g (A1) Balloon rises (B1) (Floats when) mass/weight of balloon less than mass/weight of atmosphere (of same volume as balloon) (Sinks when) mass/weight of balloon greater than mass/weight of atmosphere (of same volume as balloon) (B1) Total: 7

This question in 0625/41 May/June 2017

3 A block of wood has a volume of 210 cm3 and a mass of 180 g. (a) Calculate the density of the block of wood. density = … [2] (b) The block is held just above the surface of a liquid of density 0.88 g / cm3. Predict and explain what happens when the block is released. … … … [2] [Total: 4]

4 marks

Mark scheme: 3(a) (ρ = ) m V OR 180 ÷ 210 OR 0.18 ÷ 210 C1 0.86 g / cm3 A1 3(b) floats OR words to the same effect B1 density of wood is less than density of liquid B1 Total: 4

This question in 0625/43 May/June 2017

3 All the sides of a plastic cube are 8.0 cm long. Fig. 3.1 shows the cube. 8.0 cm Fig. 3.1 (not to scale) The mass of the cube is 0.44 kg. (a) Explain what is meant by mass. … [1] (b) (i) Calculate the density of the plastic from which the cube is made. density = … [2] (ii) The density of one type of oil is 850 kg / m3. State and explain whether the cube floats or sinks when placed in a container of this oil. … … [1] (c) On the Moon, the weight of the cube is 0.70 N. (i) Calculate the gravitational field strength on the Moon. gravitational field strength = … [2] (ii) In a laboratory on the Moon, the plastic cube is held stationary, using a clamp, in a beaker of the oil of density 850 kg / m3. The arrangement is shown in Fig. 3.2. clamp cube 3.0 cm clamp stand oil bench Fig. 3.2 The lower face of the cube is 3.0 cm below the surface of the oil. Use your answer to (c)(i) to calculate the pressure due to the oil on the lower face of the cube. pressure = … [2] [Total: 8]

8 marks

Mark scheme: 3(a) (Measure of) quantity / amount of matter OR (property) that resists change in motion / speed / momentum OR measure of a body’s inertia B1 3(b)(i) d = m / V OR in words OR 0.44 / 0.0803 OR 0.44 / 5.12 × 10–4 OR 440 / 83 OR 440 / 512 OR 0.44 / 83 OR 0.44 / 512 C1 0.86 g / cm3 OR 860 kg / m3 OR 8.6 × 10–4 kg / cm3 A1 3(b)(ii) Sinks OR does not float AND (cube) denser (than oil) B1 3(c)(i) W = mg OR (g =) W / m OR 0.70 / 0.44 C1 1.6 N / kg A1 3(c)(ii) (P =) hdg OR 0.030 × 850 × 1.6 C1 41 Pa A1

This question in 0625/41 Oct/Nov 2017

1 Fig. 1.1 shows a cylinder made from copper of density 9000 kg / m3. Fig. 1.1 The volume of the cylinder is 75 cm3. (a) Calculate the mass of the cylinder. mass = … [2] (b) The gravitational field strength is 10 N / kg. (i) Calculate the weight of the cylinder. weight = … [2] (ii) State one way in which weight differs from mass. … … … [1] (c) Fig. 1.2 shows the cylinder immersed in a liquid. liquid 2.7 cm cylinder Fig. 1.2 (not to scale) The upper face of the cylinder is at a depth of 2.7 cm below the surface of the liquid. The pressure due to the liquid at the upper face of the cylinder is 560 Pa. (i) Calculate the density of the liquid. density = … [2] (ii) Explain why the cylinder does not float in this liquid. … … [1] [Total: 8]

8 marks

Mark scheme: 1(a) OR (m =) 9000 × 7.5 × 10–5 C1 (m =) 0.68 kg accept 680 g A1 1(b)(i) W = m g in any form or (W = ) m g OR (W =) 0. 68 × 10 C1 (W =) 6.8 N A1 1(b)(ii) any one of: weight has direction / mass does not weight is a vector / mass is not weight varies / mass does not mass is amount of matter weight is a force / mass is not B1 1(c)(i) ρ = h ρ g in any form OR (ρ = ) ρ / h g OR (ρ =) 560 / (0.027 × 10) C1 (ρ =) 2.1 × 103 kg / m3 A1 1(c)(ii) explains why there is a resultant downward force B1

This question in 0625/42 Oct/Nov 2017

2 Fig. 2.1 shows a measuring cylinder that contains a coloured liquid. cm3 100 90 80 70 60 50 40 h 30 20 X 10 0 Fig. 2.1 The measuring cylinder contains 82 cm3 of the liquid. The density of the liquid is 950 kg / m3. (a) Calculate the mass of the liquid. mass = … [3] (b) The height h of the liquid in the measuring cylinder is 0.094 m. (i) Calculate the pressure due to the liquid at point X in Fig. 2.1. pressure = … [2] (ii) The true pressure at point X is different from the value calculated in (b)(i). Explain why. … … [1] (c) A small object is made of steel. It is placed level with the top surface of the liquid in the measuring cylinder and then released. The object sinks in this liquid. (i) Explain why the object sinks in this liquid. … … [1] (ii) Describe how the volume of the object can now be determined. … … … [1] [Total: 8]

8 marks

Mark scheme: 2(a) C1 7.8 / 7.79 × 10N (where N is a integer) C1 0.078 / 0.0779 kg or 78 / 77.9 g A1 2(b)(i) (p = )hρ g or 0.094 × 950 × 10 C1 890 / 893 Pa A1 2(b)(ii) atmospheric pressure (is acting) B1 2(c)(i) steel is denser (than liquid) or denser than 950 kg / m3 B1 2(c)(ii) take new reading and subtract 82 (cm3) / original reading B1

This question in 0625/43 Oct/Nov 2017

3 A rectangular container has a base of dimensions 0.12 m × 0.16 m. The container is filled with a liquid. The mass of the liquid in the container is 4.8 kg. (a) Calculate (i) the weight of liquid in the container, weight = … [1] (ii) the pressure due to the liquid on the base of the container. pressure = … [2] (b) Explain why the total pressure on the base of the container is greater than the value calculated in (a)(ii). … … [1] (c) The depth of liquid in the container is 0.32 m. Calculate the density of the liquid. density = … [2] [Total: 6]

6 marks

Mark scheme: 3(a)(i) 1 3(a)(ii) (P = ) F ÷ A OR 48 ÷ (0.12 × 0.16) 1 2500 Pa 1 3(b) Atmospheric pressure (in addition to liquid pressure) 1 3(c) P = hdg or in words OR (d =) P ÷ hg OR 2500 ÷ (0.32 × 10) 1 780 kg / m3 1 OR d = M ÷ V = 4.8 ÷ (0.12 × 0.16 × 0.32) (1) 780 kg / m3 (1)

This question in 0625/41 May/June 2018

4 (a) Fig. 4.1 shows liquid in a cylinder. cylinder liquid Fig. 4.1 The depth of the liquid is 10 cm and the radius of the cylinder is 3.0 cm. The weight of the liquid in the cylinder is 2.5 N. Calculate the density of the liquid. density = … [3] (b) Fig. 4.2 shows a device that measures the pressure of a gas supply. gas supply h liquid Fig. 4.2 (i) State the name of the device. … [1] (ii) The difference h between the two liquid levels is 2.0 cm. The density of the liquid is 800 kg / m3. Calculate the difference between the pressure of the gas and atmospheric pressure. pressure difference = … [2] (iii) A similar device with a tube of smaller cross-sectional area is connected to a gas supply at the same pressure. State and explain any effect on the value of h. … … … [2] [Total: 8]

8 marks

Mark scheme: 4(a) C1 volume = (π × 0.032 × 0.1 = 2.8 × 10–4 (m3)) C1 density = (0.25 / 2.8 × 10–4) = 890 kg / m3 A1 OR mass = 250 (g) OR ρ = m / V volume = (π × 32 × 10 =) 280 cm3 density = (250 / 280 =) 0.89 g / cm3 OR ρ = F / A = hρg ρ = F / Ahg OR 2.5 / π × 0.032 × 0.1 × 10 = 890 kg / m3 4(b)(i) manometer B1 4(b)(ii) (P =) hdg OR 0.02 × 800 × 10 C1 160 Pa A1 4(b)(iii) Value of h stays the same M1 Difference in height not dependent on cross-sectional area of tube OR Pressure of a liquid column depends only on values of h, d and g A1

This question in 0625/41 Oct/Nov 2018

2 (a) Fig 2.1 shows liquid in a cylinder. cylinder liquid Fig. 2.1 Table 2.1 gives some data about the cylinder and the liquid. Table 2.1 radius of cylinder 3.5 cm weight of empty cylinder 2.5 N depth of liquid 12.0 cm density of liquid 900 kg / m3 The cylinder containing liquid is placed on a digital balance that displays the mass in kg. Calculate the reading shown on the balance. reading … kg [4] (b) Fig. 2.2 shows a device that measures the pressure of a gas. gas supply glass tube liquid 50 mm Fig. 2.2 (i) State the name of the device. … [1] (ii) The pressure of the gas is 400 Pa greater than atmospheric pressure. Calculate the density of the liquid. density = … [2] (iii) With the gas supply connected, the top of the tube on the left of the device is sealed securely with a rubber stopper. The gas pressure is then increased. State and explain what happens to the liquid in the device. … … … … [2] [Total: 9]

9 marks

Mark scheme: 2(a) C1 ρ = m / V in any form OR (m =) ρV C1 (mass = 900 × 4.62 × 10–4 = ) 0.41 (kg) A1 0.66 kg or 250 g or 0.25 kg correctly added to previous result B1 2(b)(i) manometer B1 2(b)(ii) P = ρgh in any form or (ρ =) P / gh C1 (ρ = 400 / (10 × 0.05) = ) 800 kg / m3 A1 2(b)(iii) liquid on left goes further up tube B1 pressure of gas greater than air pressure + pressure from liquid column B1

This question in 0625/42 Oct/Nov 2018

1 Fig. 1.1 is the top view of a rectangular paddling pool of constant depth. The pool is filled with sea water. 44.0 m 20.0 m Fig. 1.1 (not to scale) (a) The volume of the sea water in the pool is 264 m3. Calculate the depth of the pool. depth = … [3] (b) The mass of the sea water in the pool is 2.70 × 105 kg. Calculate the density of the sea water. Give your answer to 3 significant figures. density = … [2] (c) Calculate the pressure due to the sea water at the bottom of the pool. pressure = … [2] (d) State a suitable instrument for measuring the dimensions given in Fig. 1.1. … [1] [Total: 8]

8 marks

Mark scheme: 1(a) C1 V = A × depth in any form OR (d =) V / A C1 (d = 264 / 880 =) 0.30 m A1 1(b) ρ = m / V in any form OR (ρ =) m / V C1 (ρ = 2.7 × 105 / 264 =) 1020 kg / m3 A1 1(c) p = ρgh in any form OR (p =) ρgh C1 (p = 1020 × 10 × 0.3 =) 3 100 Pa A1 1(d) tape measure B1

This question in 0625/42 Oct/Nov 2019

1 Fig. 1.1 is the top view of a tank in an aquarium. The tank is filled with salt water. 1.6 m 1.1 m 1.0 m 3.2 m Fig. 1.1 (not to scale) The depth of the water in the tank is 2.0 m. (a) Calculate the volume of the water in the tank. volume = … [3] (b) The density of the water in the tank is 1.1 × 103 kg / m3. Calculate the mass of the water in the tank. mass = … [2] (c) Calculate the pressure due to the water at a level of 0.80 m above the base of the tank. pressure = … [3] [Total: 8]

8 marks

Mark scheme: 1(a) attempt to use 2 rectangles for A C1 A = ((1 × 3.2) + (1.1 × 1.6) = 3.2 + 1.76 =) 4.96 (m2) C1 9.9 m3 A1 1(b) ρ = m / V OR m = ρV OR (m =) 9.9 × 1.1 × 103 C1 (m =) 1.1 × 104 kg A1 1(c) depth of water = 1.2 m C1 (P =) ρgh OR (P = 1.1 × 103 × 10 × 1.2) C1 (P =) 1.3 × 104 Pa A1

This question in 0625/43 Oct/Nov 2019

3 Fig. 3.1 shows a model of a wind turbine used to demonstrate the use of wind energy to generate electricity. The wind is blowing towards the model, as shown. turbine blades circular area swept out by turbine blades wind A V Fig. 3.1 (a) The mass of air passing through the circular area swept out by the turbine blades each second is 7.5 kg. The kinetic energy of the air that passes through this circular area each second is 240 J. (i) Calculate the speed of the air. speed = … [3] (ii) The kinetic energy of the air drives a generator. State the input power of the air passing through the turbine blades. input power = … [1] (b) The output current of the generator is 2.0 A. The output potential difference (p.d.) of the generator is 11 V. (i) Calculate the output power of the generator. output power = … [2] (ii) Calculate the efficiency of the wind turbine. efficiency = … % [2] (c) The density of air is 1.3 kg / m3. Calculate the volume of air passing through the circular area swept out by the turbine blades each second. volume = … [2] [Total: 10]

10 marks

Mark scheme: 3(a)(i) KE = ½ mv2 in any form OR v2 = 2 × KE / m OR 240 = ½ × 7.5 v2 C1 v2 = 2 × 240 / 7.5 OR (v=) √{2 × 240 / 7.5 } OR (v=) √{2KE / m } C1 = 8.0 m / s A1 3(a)(ii) 240 W B1 3(b)(i) P = VI in any form OR 11 × 2 C1 22 W A1 3(b)(ii) (efficiency =) Po / Pi OR (efficiency =) Po / Pi OR (efficiency =) (11 × 2 / 240) × 100 C1 {efficiency = (11 × 2 / 240) × 100 =} 9.2 (%) A1 3(c) ρ = m / V in any form OR (V =) m / ρ OR (V = )7.5 / 1.3 C1 (V = 7.5 / 1.3 =) 5.8 m3 A1

This question in 0625/42 Feb/March 2020

2 A scientist fills a container with sea water. The container has dimensions 30 cm × 30 cm × 40 cm. The density of sea water is 1020 kg / m3. (a) Calculate the mass of the sea water in the container. mass = … [3] (b) Fig. 2.1 shows a submarine. The submarine is fully submerged in the sea. hatch top surface submarine Fig. 2.1 (i) The atmospheric pressure is 100 kPa and the total pressure on the top surface of the submarine is 500 kPa. Calculate the depth of the top surface of the submarine below the surface of the sea. depth = … [3] (ii) A hatch (an opening door) on the top surface of the submarine has an area of 0.62 m2. Calculate the downward force on the hatch due to the total pressure on the top surface of the submarine. force = … [2] [Total: 8]

8 marks

Mark scheme: 2(a) V (= 0.3 × 0.3 × 0.4) = 0.036 (m3) C1 ρ = m / V in any form OR (m =) ρV OR 1020 × 0.036 C1 (m =) 37 kg A1 2(b)(i) P = ρgh in any form C1 (h =) 400 × 103 / (1020 × 10) C1 (h =) 39 m A1 2(b)(ii) P = F / A OR (F =) PA OR 500 × 103 × 0.62 C1 (F =) 310 000 N OR 310 kN A1

This question in 0625/43 May/June 2020

1 (a) Fig. 1.1 shows a piece of glass of thickness 2.0 cm and area 0.15 m2. The density of the glass is 2.6 × 103 kg / m3. area 0.15 m2 thickness 2.0 cm Fig. 1.1 (not to scale) Calculate the weight of the piece of glass. weight = … [3] (b) The piece of glass shown in Fig. 1.1 is used as the vertical viewing window of an aquarium. The atmospheric pressure outside the aquarium is 1.0 × 105 Pa. The average pressure on the inside of the aquarium window is 1.3 × 105 Pa. Calculate the resultant force acting on the window due to these pressures and state the direction in which it acts. force = … direction of force … [4] (c) Fig. 1.2 shows a vacuum pump connected to the top of a vertical tube with its lower end immersed in a tank of liquid. The pump reduces the pressure above the column to zero and the pressure at point X is 9.6 × 104 Pa. vacuum pump point X 12 m liquid Fig. 1.2 (not to scale) Calculate the density of the liquid. density = … [3] [Total: 10]

10 marks

Mark scheme: 1(a) 78 N A3 (m=) ρV OR ρ = m / V in any form C1 W = mg C1 1(b) 4.5 × 103 N A3 (F=) (Δ)PA OR P = F / A in any form C1 (ΔP = 1.3 × 105 – 1.0 × 105 = ) 3 × 104 C1 outwards B1 1(c) (ρ =) 800 kg / m3 A3 (ρ =) P / gh OR P= ρ gh in any form C1 (ρ =) 9.6 × 104/ (10 × 12) C1

This question in 0625/42 Feb/March 2021

1 (a) Fig. 1.1 shows a sealed weather balloon which is stationary in still air. weather balloon instruments Fig. 1.1 State whether the overall density of the balloon and its instruments is greater than, less than, or the same as the density of the surrounding air. … [1] (b) At night, the gas inside the balloon cools. The pressure of the air outside the balloon remains the same. (i) State whether the balloon rises, falls or remains stationary. … [1] (ii) Explain your answer. … … … [2] (c) An object is released from the balloon. It starts at rest and eventually reaches a constant speed. (i) On the axes of Fig. 1.2, sketch a speed–time graph to show this motion. speed 0 0 time Fig. 1.2 [3] (ii) State the values of the initial acceleration and the final acceleration of the object. initial acceleration … final acceleration … [2] [Total: 9]

9 marks

Mark scheme: 1(a) same (as density of surrounding air) B1 1(b)(i) falls B1 1(b)(ii) volume decreases B1 density increases B1 1(c)(i) starts at origin B1 finishes horizontal by eye B1 gradient decreasing smoothly to 0 B1 1(c)(ii) 10 m / s2 (down) B1 0 ignore any unit B1

This question in 0625/42 May/June 2021

5 (a) An aluminium saucepan and a steel saucepan have the same dimensions. Table 5.1 shows the values of the specific heat capacity and the density of aluminium and of steel. Table 5.1 specific heat capacity density metal J / (kg °C) kg / m3 aluminium 0.91 2600 steel 0.50 7600 The mass of the aluminium saucepan is 0.41 kg. (i) Calculate the mass of the steel saucepan. mass = … [2] (ii) Calculate the thermal capacity of the aluminium saucepan. thermal capacity = … [2] (iii) Water is heated in the steel saucepan. The initial temperature of the water and the saucepan is 20 °C. Calculate the energy transfer needed to raise the temperature of the steel saucepan to 100 °C. energy = … [2] (b) Explain why metals are better thermal conductors than non-metals. … [2] [Total: 8]

8 marks

Mark scheme: 5(a)(i) 1.2 kg A2 ( ) 7600 0.41 2600 m × = volume constant so mass directly proportional to density C1 5(a)(ii) 0.37 J / °C A2 (thermal capacity =) mass × specific heat capacity C1 5(a)(iii) 48 J A2 (E =) mcΔT OR 1.2 × 0.50 × (100 – 20) in any form C1 5(b) electrons mentioned B1 (metals have) electrons free to move / delocalised (which transfer thermal energy) B1

This question in 0625/43 Oct/Nov 2021

4 (a) A sample of sand has a volume of 0.050 m3. The density of the sand is 1900 kg / m3. The specific heat capacity of the sand is 1500 J / (kg °C). (i) Calculate the mass of the sample of sand. mass = … [2] (ii) Calculate the thermal capacity of the sample of sand. thermal capacity = … [2] (iii) The initial temperature of the sample of sand is 7.0 °C. The sample of sand is heated using an electrical heater. The power of the heating element is 50 W. Calculate the time taken to increase the temperature of the sand to 19.0 °C. time = … [3] (b) In some countries, the soil is too cold for plants to grow well. In these countries, plants are grown in plastic pots and kept inside. The pots, containing soil, are placed on sand. The sand is heated using an electrical heater, as shown in Fig. 4.1. plant soil plastic sand plant pot heating element in sand Fig. 4.1 (i) Describe, in terms of molecules, how thermal energy is transferred from the heated sand through the base of the plastic pot. … … … … [2] (ii) The heating element in Fig. 4.1 remains switched on. The temperature of the sand remains constant at a value above room temperature. Explain why the temperature of the sand remains constant. … … … [2] [Total: 11]

11 marks

Mark scheme: 4(a)(i) (mass = 1900 × 0.05) = 95 kg A2 (m =) ρV in any form OR 1900 × 0.05 (C1) 4(a)(ii) (= 95 × 1500) = 140 000 J / °C / 1.4 × 105 J / °C A2 (C =) m × c (C1) Question Answer Marks 4(a)(iii) 34 000 s / 560 min / 9 h 20 min A3 Temperature rise = 12(°C) / 19–7 OR E = C × Δ θ in any form (C1) (t =) E / P in any form OR (thermal capacity × 12) / 50 (C1) 4(b)(i) sand molecules gain KE OR vibrate more OR hit (other) molecules (when heated) OR (thermal energy is transferred by) conduction B1 Energy is transferred to molecules of plastic pot in contact with sand (through collisions) OR Energy OR (lattice) vibrations transferred to neighbouring molecules B1 4(b)(ii) (sand is warmer than surroundings and so thermal) energy (constantly) is lost from the sand B1 (at a constant temperature) rate of (thermal) energy supplied to the sand is equal to rate of (thermal) energy lost from sand B1

This question in 0625/42 Feb/March 2022

5 (a) Define specific heat capacity. … … [2] (b) A bowl contains 500 cm3 of water at a temperature of 5.0 °C. The bowl of water is placed in a freezer for several hours. When the bowl is removed from the freezer, it contains ice at a temperature of –18.0 °C. The density of water is 1000 kg / m3. (i) Calculate the mass of water in the bowl when it is placed in the freezer. mass = … [2] (ii) The specific heat capacity of water is 4200 J / (kg °C). The specific heat capacity of ice is 2100 J / (kg °C). The specific latent heat of fusion of water is 3.3 × 105 J / kg. Calculate the energy given out as the water cools from 5.0 °C to ice at –18.0 °C. energy = … [5] [Total: 9]

9 marks

Mark scheme: 5(a) energy required to raise the temperature of 1 kg / 1 g / unit mass of a substance by 1 °C / unit temperature A2 energy required to raise the temperature of a substance by 1 °C C1 5(b)(i) 0.50 kg A2  = m/V in any form C1 5(b)(ii) 190 000J / 1.9  105 J / 190 kJ A5 (E=) mc∆T in any form C1 (E=) mL in any form C1 Use of c = 4200 (J / kg °C) AND ∆T = 5 C1 Use of c = 2100 AND ∆T = 18 C1

This question in 0625/43 May/June 2022

1 Fig. 1.1 shows sea water flowing down a channel into a tank without splashing. The water is flowing at a rate of 800 kg / min. The length and width of the tank are 3.10 m and 1.20 m. The density of the sea water is 1020 kg / m3. 1.20 m flowing sea water 3.10 m channel tank Fig. 1.1 (not to scale) (a) Initially, the tank is empty. Calculate the depth of water in the tank after 1.00 minute. Give your answer to three significant figures. depth = … [3] (b) The height of the water decreases by 0.420 m as it flows down the channel. Calculate the decrease in gravitational potential energy of the water each second. decrease in gravitational potential energy = … [3] (c) The water stops flowing. The depth of water in the tank is 0.800 m. Calculate the pressure at the bottom of the tank due to the water. pressure = … [3] [Total: 9]

9 marks

Mark scheme: Question Answer Marks 1(a) (depth =) 0.211 m A3 =m / V OR (V =) m / OR 800 / 1020 C1 V = l w d OR (d =) V / (l  w) OR V ÷ 3.72 C1 1(b) (∆GPE =) 56(.0) J A3 GPE = mg∆h OR (GPE =) mg∆h OR (800 / 60)  10  0.42(0) C1 (mass per second =) 800 / 60 (kg) OR their GPE per minute ÷ 60 C1 1(c) (P =) 8200 Pa A3 (P =) hg C1 (P =) 1020  10  0.8(00) (Pa) C1 OR (P =) F / A (C1) F = mg OR (C1) F = 1020  0.8(00)  3.72  10

This question in 0625/42 Oct/Nov 2022

2 Fig. 2.1 shows a ship loaded with containers. containers ship water Fig. 2.1 (a) The ship is made of steel. The density of steel is 7800 kg / m3 and the density of water is 1000 kg / m3. Explain why the ship floats in the water. … … … [2] (b) The containers with the greatest mass are loaded near the bottom of the ship. State and explain the effect on the stability of the ship of loading the containers in this way. … … … [2] (c) A crane lifts a container 48 m vertically upwards. The mass of the container is 30 000 kg. Calculate the energy transferred to the gravitational potential energy stored in the container. energy = … [2] [Total: 6]

6 marks

Mark scheme: 2(a) ship is not solid steel / there are air spaces in ship B1 (average) density of ship is less than the density of the water B1 2(b) the centre of gravity is lower and (so) the ship is more stable A2 the centre of gravity is lower OR ship more stable (C1) 2(c) 1.4  107 J OR 14 MJ OR 14 000 kJ A2 ∆Ep= mg(∆)h OR (∆Ep= ) mg(∆)h OR 30 000  9.8  48 (C1)

This question in 0625/42 Feb/March 2023

5 (a) Fig. 5.1 shows an electric heater used to heat a room. Fig. 5.1 The dimensions of the room are 4.5 m × 6.1 m × 2.4 m. The density of air is 1.2 kg / m3. (i) Show that the mass of air in the room is 79 kg. [2] (ii) The power of the heater is 1100 W. The specific heat capacity of air is 1000 J / (kg °C). Calculate the time taken to increase the temperature of the air in the room from 16.0 °C to 20.0 °C. time = … [4] (iii) Suggest one reason why the time calculated in (a)(ii) is the minimum time needed to increase the temperature of the air in the room from 16.0 °C to 20.0 °C. … … [1] (b) Fig. 5.2 shows a cross-section of a double-glazed window in the room. outer glass pane narrow air gap inner glass pane Fig. 5.2 State the main methods of thermal energy transfer from the room to outside which are reduced by this type of window. … [1] [Total: 8]

8 marks

Mark scheme: 5(a)(i) B1 (m =) 1.2  4.5  6.1  2.4 (= 79 kg) OR (m =) 79.056 (kg) B1 5(a)(ii) 290 s A4 c = (∆)E / m∆ OR (∆E =) mc∆ OR (∆E =) 79  1000  4(.0) OR (∆E =) 316 000 OR (∆ =) 4(.0) C1 P = (∆)E / t OR (∆E =) Pt OR (∆E =) 1100  t C1 (t =) mc∆ / P OR (t =) 79  1000  4(.0) / 1100 OR (t =) 316 000 / 1100 C1 5(a)(iii) any one from:  (thermal) energy is transferred to furniture / walls / objects (in the room)  (thermal) energy is transferred through windows / doors / floor / ceiling / from the room B1 5(b) conduction AND convection B1

This question in 0625/42 May/June 2023

9 Fig. 9.1 shows the Sun as the central dot and the planets Saturn, Jupiter and Earth labelled S0, J0 and E0. The planets orbit the Sun anticlockwise. From the Earth’s orbit, the planets appear aligned. S0 J0 E0 Fig. 9.1 (not to scale) Assume that Saturn takes 30 years to orbit the Sun and that Jupiter takes 12 years to orbit the Sun. (a) On Fig. 9.1, mark the positions of Saturn and Jupiter 5.0 years after the original positions shown. Label these positions S1 and J1. Show your working. [3] (b) (i) On Fig. 9.1, mark the positions of Saturn and Jupiter 20 years after the original positions shown in Fig. 9.1. Label these positions S2 and J2. [1] (ii) State what is observed from the Earth’s orbit after 20 years. … … [1] (c) (i) Choose two words from the list to describe each planet. gaseous large rocky small Jupiter … Earth … [1] (ii) The average density of Jupiter is much less than that of the Earth. The gravitational field strength at the surface of Jupiter is greater than that at the surface of the Earth. Explain how these differences in density and in gravitational field strength are consistent with your answers to (c)(i). density … … gravitational field strength … … [3] (d) The average density of Jupiter is 1300 kg / m3 and its volume is 1.4 × 1015 km3. Calculate the mass of Jupiter. mass = … [3] [Total: 12]

12 marks

Mark scheme: 9(a) both positions correctly marked S1 and J1 on the diagram A3 Saturn moved 360  5 / 30 (= 60°) OR S1 in correct position C1 Jupiter moved 360  5 / 12 (= 150°) OR J1 in correct position C1 9(b)(i) S2 at 240° AND J2 at (600° – 360° =) 240° B1 9(b)(ii) (Saturn and Jupiter) are aligned / Jupiter exactly in front of Saturn / there is a conjunction owtte B1 9(c)(i) (Jupiter) gaseous AND large AND (Earth) rocky AND small B1 9(c)(ii) any three from: (density)  Jupiter has a low density because it is composed of gas / Earth has a high density because it is a solid (gravitational field strength)  Jupiter (has a large GFS so it) has a large mass / Earth (has a small GFS so it) has a small mass  Jupiter’s mass is larger than the Earth’s mass because the volume of Jupiter is larger even though the density of Jupiter is smaller B3 9(d) (mass =) 1.8  1027 kg A3 ρ = m / V OR (m =) ρV OR (m =) 1300  1.4  1015  109 C1 (m =) 1300  1.4  1015  109 (kg) OR 1.8(2)  10N C1

This question in 0625/43 May/June 2023

1 (a) Oil of density 0.80 g / cm3 is poured gently onto the surface of water of density 1.0 g / cm3. The oil and the water do not mix. Describe and explain the final position of the oil relative to the water. description … … explanation … … [2] (b) An irregularly shaped solid object has a density of 2.7 g / cm3. (i) Describe a method to measure the volume of the irregularly shaped solid object. … … … [2] (ii) The volume of the object is 83 cm3. Calculate the mass of the object. mass = … [3] [Total: 7]

7 marks

Mark scheme: 1(a) (oil) stays on surface / floats on the water B1 (oil has) lower density than water B1 OR liquids of lower density float on liquids of higher density 1(b)(i) measure (initial) volume of liquid / water AND immerse object B1 OR immerse object in a known / measured volume of liquid / water subtract initial / start volume from final / new volume B1 OR calculate the difference in volume OR measure change in volume OR (alternative answer) fill displacement can / container with water AND immerse object (B1) measure volume of displaced water (B1) 1(b)(ii) 220 g A3 ρ = m / V OR (m =) ρV OR 2.7  83 C1 (m =) 2.7  83 OR 2.2  10n C1 Question Answer Marks

This question in 0625/43 Oct/Nov 2023

4 (a) Define specific heat capacity. … … … [2] (b) A volume of 0.0024 m3 of oil is heated in a pan for 7.0 min. The temperature of the oil increases from 20 °C to 180 °C. The density of the oil is 910 kg / m3. The specific heat capacity of the oil is 2000 J / (kg °C). (i) Calculate the mass of oil in the pan. mass = … [2] (ii) Calculate the energy required to increase the temperature of the oil. energy = … [2] (iii) Calculate the power required to supply the energy calculated in (b)(ii). power = … [2] [Total: 8]

8 marks

Mark scheme: 4(a) energy transferred per unit mass per unit temperature change A2 (thermal) energy (transferred) per unit temperature change (C1) 4(b)(i) (m =) 2.2 kg A2 m = V in any form OR 910  0.0024 (C1) 4(b)(ii) 7.0  105 J OR 700 000 J A2 c = ∆E / { m∆} OR (∆E =) mc∆OR 2.2  2000  160 OR 2.184  2000  160 (C1) 4(b)(iii) 1700 W A2 (P =) E / t OR 700 000 / 7  60 OR 704 000 / 7  60 OR 698 880 / 7  60 (C1)

This question in 0625/42 Feb/March 2024

4 Fig. 4.1 shows a stainless-steel saucepan being heated on an electric cooker. The saucepan contains water. Fig. 4.1 (a) State what happens to the water particles as the water temperature increases. … [1] (b) The saucepan contains 250 cm3 of water. The specific heat capacity of water is 4200 J / (kg °C). The density of water is 1000 kg / m3. (i) Show that the mass of the water in the saucepan is 0.25 kg. [2] (ii) Calculate the energy required to increase the water temperature from 20 °C to 65 °C. energy = … [3] (iii) The heater supplies enough power to heat the water in 39 s. A student measures the time taken to heat the water as 115 s. Suggest why the actual time taken to heat the water is longer. Assume that the student takes accurate measurements. … … [1] (c) The stainless-steel saucepan is replaced with an aluminium saucepan of the same mass. It contains the same volume of water. The specific heat capacity of stainless steel is 500 J / (kg °C). The specific heat capacity of aluminium is 890 J / (kg °C). Explain how using an aluminium saucepan will affect the time taken to heat the water. … … … [2] [Total: 9]

9 marks

Mark scheme: 4(a) (average) KE of particles increases / particles move faster B1 4(b)(i)  = m / v OR (m =) v M1 1 cm3 = 1  10–6 m3 OR 250 cm3 = 2.5  10–4 m3 OR 1000  2.5  10–4 (= 0.25 kg) A1 4(b)(ii) 47000 J A3 ( =) 65 – 20 °C OR ( =) 45 °C C1 E = mc OR (E =) mc OR (E =) 0.25  4200  45 C1 4(b)(iii) thermal energy also transferred to the pan / surroundings OR thermal energy escapes from the water (as it is being heated) B1 4(c) any two from:  (aluminium saucepan) takes longer to heat the water  more (thermal) energy is needed (with aluminium pan for the same increase in temperature)  (because aluminium) has a higher specific heat capacity B2

This question in 0625/43 May/June 2024

2 Fig. 2.1 shows a balanced, uniform metre ruler made of wood. metre ruler 0 cm 10 cm 42 cm 80 cm 100 cm 6.0 × 10–3 m 2.6 × 10–2 m pivot 0.34 N 0.12 N Fig. 2.1 The width of the metre ruler is 2.6 × 10–2 m and the thickness of the ruler is 6.0 × 10–3 m. (a) Define the ‘moment’ of a force in words. … … [1] (b) On Fig. 2.1, mark the position of the centre of gravity of the metre ruler with a point labelled X. Label the distance of X from the 0 cm end of the ruler. [1] (c) (i) Show that the mass of the metre ruler is 0.081 kg. [3] (ii) Calculate the density of the wood of the metre ruler. density = … [2] [Total: 7]

7 marks

Mark scheme: 2(a) (moment =) force  perpendicular distance (from the pivot) B1 2(b) X AND 50 cm labelled to the right of the pivot AND left of 80 cm B1 2(c)(i) sum of clockwise moments = sum of anticlockwise moments B1 {mg  8} + {0.12  38} = {0.34  32} B1 any one from: B1 • mg = [{0.34  32} – {0.12  38}] ÷ 8 • mg = {10.88 – 4.56} ÷ 8 • mg = 6.32 ÷ 8 • mg = 0.79 • (mass =) 0.79 ÷ 9.8 2(c)(ii) 520 kg / m3 A2 (density =) mass ÷ volume OR 0.081 ÷ {1(.0)  2.6  10–2  6(.0)  10–3} C1

This question in 0625/42 May/June 2025

2 Table 2.1 contains information about the planet Mars. Table 2.1 mass 6.4 × 1023 kg gravitational field strength 3.7 N / kg at surface average density 3900 kg / m3 (a) Define gravitational field strength. … … [1] (b) (i) An object has a weight of 42 N at the surface of the Earth. Calculate the weight of the object at the surface of Mars. weight = … [2] (ii) Calculate the volume of Mars. volume = … [2] (c) Fig. 2.1 shows a space buggy that is tested on Earth. The buggy is travelling at a constant speed in a straight line. The driving force on the buggy is 30 N. 30N Fig. 2.1 (i) Draw and label one arrow on Fig. 2.1 to show the size and direction of the resistive forces on the buggy. [2] (ii) Air resistance on Mars is less than air resistance on Earth. The same driving force, 30 N, is exerted on the buggy on Mars. 1. State the effect this has on the resultant force on the buggy on Mars. … 2. State the relationship between resistive forces, driving force and resultant force. … [1] [Total: 8]

8 marks

Mark scheme: 2(a) (gravitational field strength is) force per unit mass (on an object in a gravitational field) B1 2(b)(i) 16 N A2 W = mg OR (m =) W÷g OR 42 / 9.8 OR (mass of object =) 4.3 (kg) C1 2(b)(ii) 1.6  1020 m3 A2 (V =) m / ρ OR (V =) 6.4  1023 / 3900 C1 2(c)(i) arrow parallel to driving force AND pointing to the right B1 2(c)(i) (arrow pointing to the right) labelled 30 N B1 2(c)(ii) 1 (resultant force) increases OR there is a resultant force (in the direction of the driving force) B1 OR 2 resultant force = driving force – resistive force(s)

This question in 0625/41 Oct/Nov 2025