AS & A Level Physics (9702)
Temperature Banker
28 questions · about 2% of the marks we indexed · examined in 9 of 9 sessions.
Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Specific heat capacity & specific latent heat · Core · Paper 4 style · 4 marksAn electric kettle of power 2.2 kW contains 1.5 kg of water at 22 °C. The specific heat capacity of water is 4200 J kg⁻¹ K⁻¹.
- (a) Define specific heat capacity. [1]
- (b) Calculate the energy needed to heat the water to 100 °C. [2]
- (c) Calculate the minimum time taken, assuming no energy is lost. [1]
Show answer
- (a) The energy needed per unit mass to raise the temperature of a substance by one kelvin (1 °C).
B1
- (b) 4.91 × 10⁵ J
C1 Q = mcΔθ = 1.5 × 4200 × 78; A1 4.91 × 10⁵ J
- (c) 223 s
t = Q/P = 223 s (A1)
- (a) Define specific latent heat of vaporisation. [2]
- (b) Calculate the energy needed to turn 0.25 kg of water at 100 °C into steam at 100 °C. The specific latent heat of vaporisation of water is 2.26 × 10⁶ J kg⁻¹. [1]
Show answer
- (a) The energy needed per unit mass to change a substance from liquid to vapour (gas) without any change of temperature.
B1 energy per unit mass, liquid to gas; B1 at constant temperature
- (b) 5.65 × 10⁵ J
Q = mL (A1)
- (a) Explain, in terms of molecules, why the temperature of a pure substance stays constant while it melts, even though energy is being supplied. [2]
- (b) Explain why the specific latent heat of vaporisation of a substance is much greater than its specific latent heat of fusion. [2]
Show answer
- (a) The energy supplied increases the potential energy of the molecules (breaking bonds / increasing separation), not their kinetic energy; temperature depends on the mean kinetic energy, so it stays constant.
B1 energy increases potential energy of molecules; B1 mean kinetic energy (temperature) unchanged
- (b) On boiling the molecules are separated completely (to large distances), needing much more energy than the small increase in separation on melting; work is also done pushing back the atmosphere as the volume increases greatly.
B1 much greater increase in separation / all bonds broken; B1 work done against atmosphere on expansion
Ice cubes of total mass 0.080 kg at 0 °C are taken from a freezer. The specific latent heat of fusion of ice is 3.34 × 10⁵ J kg⁻¹. The specific heat capacity of water is 4200 J kg⁻¹ K⁻¹.
- (a) Calculate the energy needed to melt the ice and then warm the meltwater to 15 °C. [3]
- (b) Suggest why ice at 0 °C cools a drink more effectively than the same mass of water at 0 °C. [1]
Show answer
- (a) 31800 J
C1 mL = 0.080 × 3.34 × 10⁵ = 26700 J; C1 mcΔθ = 0.080 × 4200 × 15 = 5040 J; A1 total 31800 J
- (b) The ice must also absorb its latent heat of fusion while melting, taking much more energy from the drink.
B1
The specific heat capacity of water is 4200 J kg⁻¹ K⁻¹.
- (a) Calculate the energy needed to heat 80 kg of water in a geyser from 15 °C to 55 °C. [2]
- (b) Calculate the time taken, in minutes, using a 2.0 kW electric element, assuming no energy loss. [1]
Show answer
- (a) 1.34 × 10⁷ J
C1 Q = mcΔθ = 80 × 4200 × 40; A1 1.34 × 10⁷ J
- (b) 112 min
t = Q/P = 6720 s = 112 min (A1)
- (a) Define specific latent heat of fusion. [1]
- (b) An ice-tray containing 0.50 kg of water at 0 °C is placed in a freezer. Calculate the energy that must be removed to freeze the water completely. The specific latent heat of fusion of ice is 3.34 × 10⁵ J kg⁻¹. [1]
- (c) The freezer removes energy from the water at a rate of 60 W. Calculate the time taken, in minutes. [1]
Show answer
- (a) The energy needed per unit mass to change a substance from solid to liquid without a change of temperature.
B1
- (b) 1.67 × 10⁵ J
Q = mL (A1)
- (c) 46 min
t = 1.67 × 10⁵/60 s = 46 min (A1)
A 1.00 kg aluminium block has an electric heater and a thermometer in holes drilled in it. The heater is connected to a 12.0 V supply and the current is 3.50 A. After 300 s the temperature has risen by 12.6 K.
- (a) Calculate the energy supplied by the heater. [1]
- (b) Calculate the specific heat capacity of aluminium obtained from these results. [2]
- (c) The accepted value is about 900 J kg⁻¹ K⁻¹. Explain why the experimental value is larger, and suggest one improvement. [2]
Show answer
- (a) 12600 J
E = VIt = 12.0 × 3.50 × 300 (A1)
- (b) 1000 J kg⁻¹ K⁻¹
C1 c = E/(mΔθ); A1 1000 J kg⁻¹ K⁻¹
- (c) Some energy is lost to the surroundings, so less than the full energy raises the block's temperature; Δθ is too small and c too large. Improvement: insulate (lag) the block / start below room temperature and finish the same amount above.
B1 heat loss gives smaller Δθ so larger c; B1 improvement
A 0.20 kg copper block at 95 °C is dropped into 0.30 kg of water at 20 °C in a polystyrene cup. The specific heat capacity of copper is 385 J kg⁻¹ K⁻¹. The specific heat capacity of water is 4200 J kg⁻¹ K⁻¹.
- (a) Calculate the final temperature of the water, assuming no energy is lost to the surroundings or the cup. [3]
- (b) Explain why, at the final temperature, there is no further net transfer of energy between the copper and the water. [1]
Show answer
- (a) 24.3 °C
M1 energy lost by copper = energy gained by water; M1 0.20 × 385 × (95 − θ) = 0.30 × 4200 × (θ − 20); A1 θ = 24.3 °C
- (b) They are at the same temperature, i.e. in thermal equilibrium.
B1
A rooftop solar water heater in Lahore has a collecting area of 2.4 m². At noon it receives 650 W of solar radiation per square metre, and 55 % of this is transferred to water flowing through the panel at a steady rate of 0.012 kg s⁻¹. The specific heat capacity of water is 4200 J kg⁻¹ K⁻¹.
- (a) Calculate the useful power absorbed by the water. [1]
- (b) Calculate the rise in temperature of the water as it passes through the panel. [2]
- (c) State and explain how the temperature rise would change if the flow rate were halved, with the same absorbed power. [1]
Show answer
- (a) 858 W
650 × 2.4 × 0.55 = 858 W (A1)
- (b) 17.0 K
M1 P = (m/t) c Δθ; A1 Δθ = 858/(0.012 × 4200) = 17.0 K
- (c) It would double, because Δθ = P/((m/t)c) is inversely proportional to the mass flow rate.
B1
A 0.50 kg sample of a wax-like solid is heated by a 120 W heater; assume all the energy goes into the sample. Its temperature–time readings show: from t = 0 to t = 150 s the temperature rises steadily from 20 °C to 80 °C; from t = 150 s to t = 1050 s it stays at 80 °C; from t = 1050 s to t = 1150 s it rises steadily from 80 °C to 110 °C.
- (a) State the melting point of the substance. [1]
- (b) Calculate the specific heat capacity of the solid. [2]
- (c) Calculate the specific latent heat of fusion. [2]
- (d) Calculate the specific heat capacity of the liquid. [1]
Show answer
- (a) 80 °C
B1 temperature constant while melting
- (b) 600 J kg⁻¹ K⁻¹
C1 120 × 150 = 0.50 × c × 60; A1 600 J kg⁻¹ K⁻¹
- (c) 2.16 × 10⁵ J kg⁻¹
C1 120 × 900 = 0.50 × L; A1 2.16 × 10⁵ J kg⁻¹
- (d) 800 J kg⁻¹ K⁻¹
120 × 100/(0.50 × 30) = 800 J kg⁻¹ K⁻¹ (A1)
The specific latent heat of vaporisation of water is 2.26 × 10⁶ J kg⁻¹. The specific heat capacity of water is 4200 J kg⁻¹ K⁻¹.
- (a) Calculate the energy released when 5.0 g of steam at 100 °C condenses on the skin and the water then cools to 37 °C. [2]
- (b) Calculate the energy released by 5.0 g of boiling water cooling from 100 °C to 37 °C, and use your answers to explain why a steam scald is more serious. [2]
Show answer
- (a) 12600 J
C1 mL + mcΔθ = 0.005 × 2.26 × 10⁶ + 0.005 × 4200 × 63; A1 12600 J
- (b) 1320 J; steam releases about 9.5 times as much energy, mostly the latent heat of condensation.
B1 value; B1 comparison referring to latent heat
A car engine's cooling system must remove waste energy at a rate of 25 kW. The coolant has a specific heat capacity of 3800 J kg⁻¹ K⁻¹ and its temperature rises by 8.0 K as it passes through the engine.
- (a) Calculate the mass of coolant that must flow through the engine each second. [2]
- (b) Suggest why a liquid with a high specific heat capacity is chosen as the coolant. [1]
Show answer
- (a) 0.822 kg s⁻¹
C1 P = (m/t)cΔθ; A1 m/t = 25 000/(3800 × 8.0) = 0.822 kg s⁻¹
- (b) It absorbs a large amount of energy for a small temperature rise (so a smaller flow rate is needed).
B1
During one hour of a cricket match in hot weather in Multan, a fast bowler loses 0.80 kg of sweat, all of which evaporates from his skin. The specific latent heat of vaporisation of water at skin temperature is 2.4 × 10⁶ J kg⁻¹.
- (a) Calculate the energy removed from the body by the evaporation of the sweat. [1]
- (b) Calculate the average rate of cooling. [1]
- (c) Explain, in terms of molecules, why evaporation cools the liquid that remains. [2]
Show answer
- (a) 1.92 × 10⁶ J
Q = mL = 0.80 × 2.4 × 10⁶ (A1)
- (b) 533 W
P = 1.92 × 10⁶/3600 = 533 W (A1)
- (c) The molecules with the highest kinetic energies escape from the surface, so the mean kinetic energy of the remaining molecules falls, and temperature depends on mean kinetic energy.
B1 most energetic molecules escape; B1 mean KE of the rest falls so temperature falls
A liquid at its boiling point is evaporated by an immersed electric heater, in a flask that loses energy to the surroundings at a constant rate. With the heater at 50 W, 6.2 g of liquid evaporates in 300 s. With the heater at 80 W, 10.2 g evaporates in 300 s.
- (a) Explain why two different heater powers are used. [1]
- (b) Calculate the specific latent heat of vaporisation of the liquid. [3]
- (c) Calculate the rate of energy loss to the surroundings. [2]
Show answer
- (a) The rate of energy loss is the same in both runs (same temperature), so subtracting the two results eliminates the energy lost.
B1
- (b) 2.25 × 10⁶ J kg⁻¹
M1 P t = mL + h t for each run; M1 (80 − 50) × 300 = (10.2 − 6.2) × 10⁻³ × L; A1 L = 2.25 × 10⁶ J kg⁻¹
- (c) 3.5 W
M1 50 × 300 = 6.2 × 10⁻³ × 2.25 × 10⁶ + 300h; A1 h = 3.5 W
In a continuous-flow experiment, a liquid flows steadily through a glass tube past an electric heating wire. The inlet temperature is 21.0 °C and the outlet temperature is 23.5 °C in both runs.Run 1: p.d. 9.0 V, current 2.50 A, mass flow rate 2.0 × 10⁻³ kg s⁻¹.Run 2: p.d. 12.0 V, current 3.18 A, mass flow rate 3.5 × 10⁻³ kg s⁻¹.
- (a) Write an equation relating the electrical power VI, the mass flow rate m/t, c, the temperature rise Δθ and the rate of energy loss h. [1]
- (b) Explain why keeping the inlet and outlet temperatures the same in both runs allows h to be eliminated. [1]
- (c) Calculate the specific heat capacity of the liquid. [3]
- (d) Calculate h. [1]
Show answer
- (a) VI = (m/t)cΔθ + h
B1
- (b) The apparatus is at the same temperatures in both runs, so the rate of energy loss h is the same and cancels on subtraction.
B1
- (c) 4180 J kg⁻¹ K⁻¹
M1 subtract: 12.0 × 3.18 − 9.0 × 2.5 = (0.0035 − 0.0020) × c × 2.5; M1 15.66 = 0.00375c; A1 c = 4180 J kg⁻¹ K⁻¹
- (d) 1.6 W
h = 22.5 − 0.0020 × 4180 × 2.5 = 1.6 W (A1)
A can of cola, 0.33 kg, at 30 °C is to be cooled to 5 °C by adding ice cubes taken from a freezer at −10 °C. Treat the cola as water. Ignore energy exchange with the can and the surroundings. Specific heat capacity of ice = 2100 J kg⁻¹ K⁻¹; specific latent heat of fusion of ice = 3.34 × 10⁵ J kg⁻¹; The specific heat capacity of water is 4200 J kg⁻¹ K⁻¹.
- (a) Calculate the energy given out by the cola as it cools to 5 °C. [1]
- (b) Calculate the energy absorbed by 1.0 kg of ice warming from −10 °C to 0 °C, melting, and the meltwater warming to 5 °C. [2]
- (c) Hence calculate the mass of ice needed, in grams. [1]
Show answer
- (a) 3.47 × 10⁴ J
0.33 × 4200 × 25 = 34 650 J ≈ 3.47 × 10⁴ J (A1)
- (b) 3.76 × 10⁵ J
M1 2100 × 10 + 3.34 × 10⁵ + 4200 × 5; A1 3.76 × 10⁵ J
- (c) 92 g
m = 34600/3.76 × 10⁵ = 0.092 kg (A1)
A car of mass 1200 kg travelling at 25 m s⁻¹ on the M-2 motorway brakes to rest. Its steel brake discs have a total mass of 8.0 kg; the specific heat capacity of steel is 450 J kg⁻¹ K⁻¹.
- (a) Calculate the rise in temperature of the discs, assuming all the kinetic energy of the car is transferred to them. [3]
- (b) Suggest two reasons why the actual temperature rise is smaller. [2]
Show answer
- (a) 104 K
C1 KE = ½ × 1200 × 25² = 3.75 × 10⁵ J; M1 Δθ = KE/(mc); A1 104 K
- (b) Energy is transferred to the brake pads, tyres, road and air (and as sound); air resistance and rolling resistance also slow the car; the discs lose heat to the air while braking.
B1 each, max 2
A student investigates the melting of crushed ice at 0 °C by an immersed electric heater of power P. Meltwater drips from a funnel into a beaker; the mass m collected in t = 300 s is measured for each power. The ice also gains energy from the warm room at a constant rate.P / W | m / g20.0 | 21.730.0 | 30.440.0 | 39.750.0 | 48.460.0 | 57.5
- (a) Show that P = L(m/t) − h, where h is the rate at which energy enters the ice from the surroundings. State the gradient and y-intercept of a graph of P against m/t. [2]
- (b) Calculate m/t for each row, in mg s⁻¹. [1]
- (c) Using the first and last rows (or a best-fit line), determine L and h. [4]
- (d) Suggest why it is better to obtain L from a graph of several readings than from a single reading with h ignored. [1]
Show answer
- (a) Energy to melt per second = P + h = L(m/t), so P = L(m/t) − h; gradient = L, y-intercept = −h.
B1 equation; B1 gradient L and intercept −h
- (b) 72.3, 101.3, 132.3, 161.3, 191.7
B1 all correct (m in g × 1000/300)
- (c) L ≈ 3.35 × 10⁵ J kg⁻¹ (accept 3.1–3.6 × 10⁵); h ≈ 4.2 W (accept 2–6 W)
M1 gradient using kg s⁻¹; A1 L; M1 intercept from y = mx + c; A1 h
- (d) The graph separates the heat gained from the surroundings (intercept) from L (gradient), and averages out random errors.
B1
A 2.0 kW kettle holds 1.2 kg of water at 20 °C. It takes 230 s for the water to reach 100 °C. The kettle is then left boiling for a further 3.0 minutes, during which the mass of water falls by 0.145 kg. Specific latent heat of vaporisation of water = 2.26 × 10⁶ J kg⁻¹. The specific heat capacity of water is 4200 J kg⁻¹ K⁻¹.
- (a) Calculate the percentage of the energy supplied in the first 230 s that is used to heat the water. [2]
- (b) Calculate the average rate of energy loss from the kettle during the 3.0 min of boiling. [3]
- (c) Suggest why the percentage of energy wasted is greater during the first part of the heating than its rate of loss alone would suggest. [1]
Show answer
- (a) 88 %
C1 energy to water = 1.2 × 4200 × 80 = 4.03 × 10⁵ J, supplied = 2000 × 230 = 4.60 × 10⁵ J; A1 88 %
- (b) 179 W
M1 energy supplied = 2000 × 180 = 3.60 × 10⁵ J; M1 energy for vaporisation = 0.145 × 2.26 × 10⁶ = 3.28 × 10⁵ J; A1 loss rate = 179 W
- (c) Energy is also needed to heat the kettle's own body and element, not just the water.
B1
- (a) Explain what is meant by two objects being in thermal equilibrium. [2]
- (b) A metal spoon at 20 °C is placed in a cup of hot tea at 70 °C. State the direction of net thermal energy transfer and when it stops. [2]
- (c) State what is meant by temperature, in terms of thermal energy transfer. [1]
Show answer
- (a) There is no net transfer of thermal energy between them when they are in thermal contact, because they are at the same temperature.
B1 no net energy transfer; B1 same temperature
- (b) From the tea to the spoon; it stops when the spoon and tea reach the same temperature.
B1 tea to spoon (hot to cold); B1 same temperature
- (c) Temperature determines the direction of net thermal energy transfer: energy flows from a region of higher temperature to one of lower temperature.
B1
- (a) Convert a body temperature of 37.0 °C into kelvin. [1]
- (b) The temperature of a cup of tea falls by 25 °C. State the fall in temperature in kelvin, and explain. [1]
- (c) Liquid nitrogen boils at 77 K. Calculate this temperature in °C. [1]
Show answer
- (a) 310.2 K (310.15 K)
T/K = θ/°C + 273.15 (A1)
- (b) 25 K; one kelvin is the same size interval as one degree Celsius.
B1
- (c) −196 °C
77 − 273.15 = −196 °C (A1)
- (a) Explain why the thermodynamic (kelvin) scale is said not to depend on the property of any particular substance. [2]
- (b) State what is meant by absolute zero. [1]
Show answer
- (a) It is defined theoretically (from absolute zero and the triple point of water), whereas scales based on, for example, expansion of mercury or resistance of a wire depend on how that property varies, which differs between substances and need not be linear.
B1 thermometric property scales depend on the substance / can disagree; B1 thermodynamic scale is absolute / theoretical
- (b) The lowest possible temperature (0 K), at which a substance has minimum internal energy.
B1
- (a) State three physical properties that vary with temperature and can be used to measure temperature. [3]
- (b) State two features that make a physical property suitable for use in a thermometer. [2]
Show answer
- (a) Any three: volume of a liquid (in a glass tube); resistance of a metal wire; resistance of a thermistor; e.m.f. of a thermocouple; pressure of a gas at constant volume.
B1 each, max 3
- (b) Any two: it varies continuously (monotonically) with temperature; a measurable change for a small change in temperature (sensitive); it has a single value at each temperature; it varies over the required range; it is reproducible.
B1 each, max 2
A platinum resistance thermometer has a resistance of 100.0 Ω at the ice point (0 °C) and 138.5 Ω at the steam point (100 °C). Placed in a warm oven, its resistance is 117.3 Ω.
- (a) Calculate the temperature of the oven on the centigrade scale of this resistance thermometer. [2]
- (b) A mercury-in-glass thermometer placed beside it reads a slightly different value. Explain why, although both thermometers agree at 0 °C and 100 °C. [2]
Show answer
- (a) 44.9 °C
C1 θ = (R − R₀)/(R₁₀₀ − R₀) × 100; A1 44.9 °C
- (b) Each scale assumes its thermometric property varies linearly with temperature; resistance and mercury expansion do not vary in exactly the same (linear) way, so readings can differ between the fixed points.
B1 scales assume linear variation; B1 properties vary differently / non-linearly
A thermocouple has its cold junction in melting ice at 0 °C. When its hot junction is at 100 °C the e.m.f. is 4.10 mV. The hot junction is then touched against the tip of a soldering iron in a repair shop, and the e.m.f. becomes 2.65 mV after the reading has settled (the tip has cooled while switched off).
- (a) Assuming the e.m.f. varies linearly with temperature, calculate the temperature of the hot junction. [2]
- (b) State two advantages of a thermocouple over a liquid-in-glass thermometer for measuring the temperature of a small soldering-iron tip. [2]
Show answer
- (a) 64.6 °C
C1 θ = 2.65/4.10 × 100; A1 64.6 °C
- (b) Any two: very small junction, so small heat capacity and does not cool the object; fast response; can measure the temperature at a point; wide range (high temperatures); electrical output can be recorded remotely.
B1 each, max 2
In a constant-volume gas thermometer the pressure of the gas is proportional to its thermodynamic temperature. The pressure is 52.0 kPa at the triple point of water (273.16 K) and 71.5 kPa when the bulb is placed in a hot liquid.
- (a) Calculate the thermodynamic temperature of the liquid. [2]
- (b) Convert this temperature to degrees Celsius. [1]
Show answer
- (a) 375.6 K
C1 T = 273.16 × p/p_tr; A1 375.6 K
- (b) 102.4 °C
θ = T − 273.15 (A1)
- (a) A student says that heating air from 20 °C to 40 °C doubles its temperature. Explain why this is wrong. [1]
- (b) Calculate the Celsius temperature at which the thermodynamic temperature of the air would be double its value at 20 °C. [2]
Show answer
- (a) Temperature ratios are only meaningful on the thermodynamic scale: 293 K to 313 K is an increase of only about 7 %.
B1
- (b) 313 °C
C1 2 × 293.15 = 586.3 K; A1 586.3 − 273.15 = 313 °C
The resistance R of a platinum wire varies with Celsius temperature θ as R = R₀(1 + αθ + βθ²), where R₀ = 100.0 Ω, α = 3.9 × 10⁻³ K⁻¹ and β = −5.8 × 10⁻⁷ K⁻².
- (a) Calculate the resistance of the wire at 100 °C. [1]
- (b) The resistance in a furnace is 150.0 Ω. Calculate the temperature given by a scale that assumes R varies linearly between 0 °C and 100 °C. [2]
- (c) Calculate the actual Celsius temperature of the furnace. [3]
- (d) Explain why the two values differ. [1]
Show answer
- (a) 138.42 Ω
R = 100(1 + 0.39 − 0.0058) (A1)
- (b) 130.1 °C
C1 θ = (150.0 − 100.0)/(138.42 − 100.0) × 100; A1 130.1 °C
- (c) 130.7 °C
M1 3.9 × 10⁻³θ − 5.8 × 10⁻⁷θ² = 0.500; M1 quadratic formula (smaller root); A1 130.7 °C
- (d) The resistance does not vary linearly with temperature (β ≠ 0), but the centigrade scale assumes it does.
B1
A thermometer of heat capacity 12 J K⁻¹, initially at 22 °C, is placed in 25 g of water at 60 °C. The specific heat capacity of water is 4200 J kg⁻¹ K⁻¹.
- (a) Calculate the temperature shown by the thermometer once thermal equilibrium is reached. Ignore energy exchange with the surroundings. [3]
- (b) Suggest how the thermometer should be chosen to reduce this error. [1]
Show answer
- (a) 56.1 °C
M1 energy gained by thermometer = energy lost by water; M1 12(θ − 22) = 0.025 × 4200 × (60 − θ); A1 θ = 56.1 °C
- (b) Use a thermometer with a much smaller heat capacity (e.g. a thermocouple or small thermistor), or a much larger sample of water.
B1
A student calibrates an NTC thermistor for use as a thermometer. It is suggested that its resistance R is related to temperature by R = A e^(B/T), where A and B are constants. She records R at several Celsius temperatures θ using a water bath.θ / °C | R / Ω20.0 | 1234035.0 | 676050.0 | 403065.0 | 252080.0 | 1590
- (a) A graph of ln R against 1/T is plotted, where T is the thermodynamic temperature. Express the gradient and y-intercept in terms of A and B. [1]
- (b) Calculate 1/T (in 10⁻³ K⁻¹) and ln R for each row. [2]
- (c) Using the first and last rows (or a best-fit line), determine B. [2]
- (d) Use your value of B and the reading at 50.0 °C to estimate the resistance at 0 °C. [2]
Show answer
- (a) gradient = B; y-intercept = ln A
B1
- (b) 3.411, 9.42; 3.245, 8.82; 3.095, 8.30; 2.957, 7.83; 2.832, 7.37
B1 T = θ + 273.15 used; B1 ln R values to 2 or more d.p.
- (c) B ≈ 3540 K (accept 3300–3700 K)
M1 gradient using 1/T in K⁻¹; A1 B in range with unit K
- (d) ≈ 2.99 × 10⁴ Ω (accept 2.9–3.1 × 10⁴ Ω, consistent with B)
M1 R₀/R₅₀ = exp(B(1/273.15 − 1/323.15)); A1 answer consistent with B
