Questions on internal energy and changes of state

Question 1

Define the internal energy of a substance.

Accepted Answer

The internal energy is the sum of the kinetic and potential energies of all the particles in the substance (1). These energies are randomly distributed (1).

Question 2

Calculate the energy needed to raise the temperature of 0.50 kg of water by 20°C (specific heat capacity of water = 4200 J kg⁻¹ K⁻¹).

Accepted Answer

Q = mcΔθ = 0.50 × 4200 × 20 (1) Q = 42000 J (1)

Question 3

Explain why the temperature of a substance stays constant while it changes state, even though energy is being supplied.

Accepted Answer

The supplied energy goes into breaking the bonds between particles, increasing their potential energy, not into increasing their kinetic energy (1). Since temperature depends on average kinetic energy, it stays constant until the change of state is complete (1).

Question 4

State what is meant by two objects being in thermal equilibrium.

Accepted Answer

They are at the same temperature, so there is no net transfer of energy between them (1).

Question 5

The gas in a sealed cylinder gains 250 J of energy by heating. At the same time a piston does 90 J of work on the gas. Calculate the increase in the internal energy of the gas.

Accepted Answer

ΔU = q + w (1) ΔU = 250 + 90 = 340 J (1)

Question 6

A baking tray warms from 18°C to 43°C in an oven. State the temperature rise of the tray in kelvin, and explain why no unit conversion is needed.

Accepted Answer

Δθ = 43 − 18 = 25 K (1). A temperature difference has the same numerical value in °C and K, because the two scales have intervals of equal size and differ only in the position of their zero (1).

Question 7

An aluminium block of mass 2.0 kg is heated from 20°C to 70°C. Calculate the energy supplied to the block (specific heat capacity of aluminium = 900 J kg⁻¹ K⁻¹). Assume no energy is lost to the surroundings.

Accepted Answer

Δθ = 70 − 20 = 50 K (1) Q = mcΔθ = 2.0 × 900 × 50 (1) Q = 90000 J (1)

Question 8

Calculate the energy needed to melt 0.30 kg of ice at its melting point (specific latent heat of fusion = 3.34 × 10⁵ J kg⁻¹), and state why the temperature does not rise while the ice melts.

Accepted Answer

Q = ml = 0.30 × 3.34 × 10⁵ (1)
Q = 1.00 × 10⁵ J (1)
The energy goes into breaking bonds between molecules, raising their potential energy (1)
It does not raise their kinetic energy, so the temperature stays constant (1)

Question 9

Supplying 5000 J of energy raises the temperature of a 0.20 kg block by 25°C. Calculate the specific heat capacity of the block.

Accepted Answer

c = Q/(mΔθ) (1) c = 5000/(0.20 × 25) (1) c = 1000 J kg⁻¹ K⁻¹ (1)

Question 10

A 2.0 kW heater warms 1.5 kg of water by 30°C (c = 4200 J kg⁻¹ K⁻¹). Assuming no losses, calculate the time taken.

Accepted Answer

Q = mcΔθ = 1.5 × 4200 × 30 = 189000 J (1) t = Q/P = 189000/2000 (1) t = 94.5 s (1)

Question 11

In a continuous-flow experiment, water flows at 8.0 g s⁻¹ over a 1.50 kW electric heater. The water enters at 18.0°C and leaves at 58.0°C (c_water = 4200 J kg⁻¹ K⁻¹). Calculate the rate at which energy is lost to the surroundings.

Accepted Answer

Each second 8.0 g = 0.0080 kg of water flows through (1) Energy gained by the water each second = mcΔθ = 0.0080 × 4200 × 40.0 = 1344 W (1) Rate of loss = 1500 − 1344 = 156 W (1)

Question 12

A camping stove transfers energy to a pan of water at 750 W. The pan holds 0.90 kg of water at 15°C (c = 4200 J kg⁻¹ K⁻¹). Show that the water takes about 7 minutes to reach 100°C. Assume no energy is lost and that the heat capacity of the pan itself is negligible.

Accepted Answer

Q = mcΔθ = 0.90 × 4200 × 85 = 321300 J (1) t = Q/P = 321300/750 (1) t = 428 s = 7.1 min, which is about 7 minutes (1)

Question 13

An overnight storage tank holds 140 kg of water, which an immersion heater warms from 12°C to 55°C (c = 4200 J kg⁻¹ K⁻¹). Electricity costs 28p per kWh (1 kWh = 3.6 × 10⁶ J). Determine the cost of heating the tank. Assume no energy is lost.

Accepted Answer

Q = mcΔθ (1) Q = 140 × 4200 × 43 = 2.528 × 10⁷ J (1) E = 2.528 × 10⁷/(3.6 × 10⁶) = 7.02 kWh (1) Cost = 7.02 × 28p = 197p ≈ £1.97 (1)

Question 14

The simple kinetic model describes a state of matter by three properties of its particles: their spacing, their ordering and their motion. Describe a solid, a liquid and a gas in terms of all three of these properties. Two of the six marks are available for each property.

Accepted Answer

Spacing: in a solid and in a liquid the particles are touching their neighbours, and melting barely changes the separation (1)
Spacing: in a gas the particles sit about ten times further apart than in a liquid, roughly a thousand times the volume for the same number of particles (1)
Ordering: a solid's particles sit in a regular repeating arrangement, held there by strong forces between them (1)
Ordering: a liquid is disordered beyond its nearest few neighbours, and a gas is in no arrangement whatsoever (1)
Motion: each particle of a solid vibrates about a fixed position and cannot swap places with the particles around it (1)
Motion: a liquid's particles slide past one another while staying in contact, and a gas's travel in rapid straight lines in random directions, feeling almost no force from one another between collisions (1)

Question 15

Calculate the total energy needed to turn 0.20 kg of ice at 0°C into water and then warm it to 20°C (latent heat of fusion = 3.34 × 10⁵ J kg⁻¹, c_water = 4200 J kg⁻¹ K⁻¹).

Accepted Answer

Melting: Q = ml = 0.20 × 3.34 × 10⁵ (1) Q = 66800 J (1) Warming: Q = mcΔθ = 0.20 × 4200 × 20 = 16800 J (1) Total = 83600 J (1)

Question 16

0.10 kg of water at 80°C is mixed with 0.20 kg of water at 20°C. Assuming no energy is lost to the surroundings, calculate the final temperature.

Accepted Answer

Energy lost by the hot water = energy gained by the cold water (1) 0.10 × c × (80 − T) = 0.20 × c × (T − 20) (1) The c cancels (1) T = 40°C (1)

Question 17

Explain the difference between heating and doing work as ways of increasing the internal energy of a gas.

Accepted Answer

Heating transfers energy because of a temperature difference, from a hotter body to a cooler one (1). Doing work transfers energy mechanically, for example by compressing the gas with a piston (1). Both raise the internal energy, but by different means (1).

Question 18

A manufacturer of storage heaters needs a solid core that stores 5.4 MJ of energy when its temperature rises from 20°C to 220°C. The core must fit inside a casing of internal volume 0.0090 m³. Three materials are available. Material W: c = 880 J kg⁻¹ K⁻¹, density 2300 kg m⁻³, cost 15p per kg. Material X: c = 450 J kg⁻¹ K⁻¹, density 7200 kg m⁻³, cost 60p per kg. Material Y: c = 930 J kg⁻¹ K⁻¹, density 5100 kg m⁻³, cost 90p per kg. Deduce which material the manufacturer should choose.

Accepted Answer

Mass needed m = Q/(cΔθ): W needs 30.7 kg, X needs 60.0 kg, Y needs 29.0 kg (1)
Volume = m/ρ: W 0.0133 m³, X 0.0083 m³, Y 0.0057 m³ (1)
W is larger than the 0.0090 m³ casing, so it is rejected (1)
Cost: X = 60.0 × 60p = £36, Y = 29.0 × 90p = £26 (1)
Both X and Y fit, and Y is cheaper, so the manufacturer should choose Y (1)

Question 19

A 2.4 kW kettle contains 0.50 kg of water at 100°C. The kettle is faulty and does not switch off (specific latent heat of vaporisation of water = 2.26 × 10⁶ J kg⁻¹). Show that the kettle boils dry in about 8 minutes. Assume all the energy supplied goes to the water.

Accepted Answer

Q = ml = 0.50 × 2.26 × 10⁶ = 1.13 × 10⁶ J (1) t = Q/P = 1.13 × 10⁶/2400 (1) t = 471 s = 7.8 min, which is about 8 minutes (1)

Question 20

A student heats a beaker of crushed ice, initially at −15°C, using an immersion heater that supplies energy at a constant rate, and continues until the melted water has boiled for some time. Describe and explain the shape of the temperature-time graph for the contents of the beaker. Refer in your answer to the kinetic and potential energies of the molecules and to the relative sizes of the quantities involved.

Accepted Answer

The graph shows a rising section for the ice, a plateau at 0°C, a rising section for the water and a second plateau at 100°C (1). During the rising sections the energy supplied increases the average kinetic energy of the molecules, and temperature measures this kinetic energy (1). During the plateaus the energy supplied increases the potential energy of the molecules as intermolecular bonds are broken, so the kinetic energy and the temperature are unchanged (1). The two rising sections have different gradients because ice and water have different specific heat capacities; the smaller c of ice gives the steeper line for the same power (1). The plateau at 100°C is much longer than the plateau at 0°C because the specific latent heat of vaporisation is much larger than the specific latent heat of fusion (1). Because the power is constant, the time spent on each section of the graph is proportional to the energy that stage needs (1).

Question 21

In a continuous-flow experiment a liquid passes a heater at 12.0 g s⁻¹ while the heater delivers 1290 W, and the liquid leaves 25.0 K warmer than it entered. The flow rate is then changed to 8.0 g s⁻¹ and the power adjusted to 870 W, giving the same inlet and outlet temperatures as before. Determine the specific heat capacity of the liquid and the rate at which energy is lost to the surroundings, and explain why a second run is needed.

Accepted Answer

Each run gives P = mcΔθ + h, with m the mass passing per second and h the rate of energy loss: 1290 = 0.0120cΔθ + h and 870 = 0.0080cΔθ + h (1)
The inlet and outlet temperatures are the same in both runs, so h is the same and subtracting eliminates it: P₁ − P₂ = (m₁ − m₂)cΔθ (1)
c = (1290 − 870)/((0.0120 − 0.0080) × 25.0) (1)
c = 4200 J kg⁻¹ K⁻¹ (1)
h = 1290 − 0.0120 × 4200 × 25.0 = 30 W; a single run has both c and h unknown, so one run cannot give either (1)

Question 22

An electrical method is used to determine the specific latent heat of fusion of ice and, in a separate apparatus, the specific latent heat of vaporisation of a liquid. For the fusion determination, state what a second, identical funnel of ice with its heater switched off is used to find, and how its meltwater is used in the result. For the vaporisation determination, a heater in the liquid, once it is boiling steadily, delivers 90 W and the balance shows 15 g of liquid lost in 120 s; raised to 150 W, with the liquid boiling just as steadily, it loses 35 g in the same 120 s. Calculate the specific latent heat of vaporisation of the liquid, and explain why subtracting the two runs removes the heat lost to the surroundings from the result.

Accepted Answer

The second funnel, with its heater switched off, measures the mass the room alone melts in the same time, m₀ (1)
The heater's own meltwater m is then credited only with the excess over this: VIt = (m − m₀)l_f (1)
l_v = (P₂ − P₁)t/(m₂ − m₁) = (150 − 90) × 120/(0.035 − 0.015) (1)
l_v = 3.6 × 10⁵ J kg⁻¹ (1)
The liquid is boiling just as steadily, at the same temperature, in both runs, so the rate of heat loss to the surroundings, h, is the same number in each; subtracting the two power equations cancels h exactly without it ever being measured, leaving only the extra power that boiled away the extra mass (1)

Question 23

A metal block of mass 1.00 kg is heated by an immersion heater delivering a constant 45.0 W, and its temperature is recorded every 60 s. Determine the specific heat capacity of the metal from the readings, giving its unit, and explain whether energy lost to the room makes this value larger or smaller than the true one.

Accepted Answer

Rate of temperature rise taken between two widely separated readings, (34.0 − 19.0)/(300 − 0) (1)
= 0.0500 K s⁻¹, the readings rising by a steady 3.0 K every 60 s (1)
Applying Q = mcΔθ to one second of heating, P = mc × (rate), so c = P/(m × rate) (1)
c = 45.0/(1.00 × 0.0500) = 900 J kg⁻¹ K⁻¹, unit required (1)
The block loses energy to the room, so less than 45.0 W actually raises its temperature; crediting the whole 45.0 W to the block makes the calculated value of c larger than the true one (1)

time t / s	temperature θ / °C
0	19.0
60	22.0
120	25.0
180	28.0
240	31.0
300	34.0