CAIE A-Level Physics 14 Temperature

Practise analysing thermal equilibrium and temperature scales, evaluating thermometers and calculating heating, phase-change and energy-transfer quantities.

Exam points

analyse temperature measurement, thermal equilibrium and energy transfer
convert and interpret thermodynamic temperature scales, including absolute zero
calculate energy changes using specific heat capacity and specific latent heat
distinguish sensible heating from phase-change energy and interpret experimental data

Covered syllabus points

14.1 Thermal equilibrium 14.2 Temperature scales 14.3 Specific heat capacity and specific latent heat

[Maximum number: 10]

Use the information in (i) to draw, on Fig. 1.1, a line to represent the temperature of the block, assuming no energy losses to the surroundings.

(a)

State

[ 2 ]

(i)

what may be deduced from the difference in the temperatures of two objects,

[ 1 ]

(ii)

the basic principle by which temperature is measured.

[ 1 ]

(b)

By reference to your answer in (a)(ii), explain why two thermometers may not give the same temperature reading for an object.

[ 2 ]

(c)

A block of aluminium of mass 670 g is heated at a constant rate of 95 W for 6.0 minutes. The specific heat capacity of aluminium is $910 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$ . The initial temperature of the block is $24^{\circ} \mathrm{C}$ .

[ 6 ]

(i)

Assuming that no thermal energy is lost to the surroundings, show that the final temperature of the block is $80^{\circ} \mathrm{C}$ .

[ 3 ]

$P t=m c(\Delta) \theta$

C1

$95 \times 6 \times 60=0.670 \times 910 \times \Delta \theta$

M1

$\Delta \theta=56^{\circ} \mathrm{C}$ so final temperature $=56+24=80^{\circ} \mathrm{C}$

A1

or

$95 \times 6 \times 60=0.67 \times 910 \times(\theta-24)$

(M1)

so final temperature or $\theta=80^{\circ} \mathrm{C}$

(A1)

(ii)

In practice, there are energy losses to the surroundings. The actual variation with time t of the temperature $\theta$ of the block is shown in Fig. 1.1.

Fig. 1.1

[ 3 ]

1. sketch: straight line from (0,24) to (6,80)

2. temperature drop due to energy loss $=(80-64)=16^{\circ} \mathrm{C}$

energy loss $=0.670 \times 910 \times(80-64)=9800\,\mathrm{J}$

or

energy to raise temperature to $64^{\circ} \mathrm{C}=0.670 \times 910 \times(64-24)$ $=24400\,\mathrm{J}$ $\text{loss}=(95 \times 6 \times 60)-24400=9800\,\mathrm{J}$

[Maximum number: 8]

Fig. 2.1 shows a laboratory thermometer that is calibrated to measure temperature in degrees Celsius.

Fig. 2.1

The thermometer makes use of the fact that the density of mercury varies with temperature.

(a)

State two other physical properties of materials, apart from the density of a liquid, that can be used for measuring temperature.

1

2

[ 2 ]

(b)

The thermometer is initially at $23.0^{\circ} \mathrm{C}$ , as shown in Fig. 2.1. It is used to measure the temperature of an insulated beaker of water that is at $37.4^{\circ} \mathrm{C}$ . The bulb of the thermometer is inserted into the water, and the water is stirred until the reading on the thermometer becomes steady.

The mass of water in the beaker is 18.7 g .
The mass of mercury in the thermometer is 6.94 g .
The specific heat capacity of water is $4.18 \mathrm{~J} \mathrm{~g}^{-1} \mathrm{~K}^{-1}$ .
The specific heat capacity of mercury is $0.140 \mathrm{~J} \mathrm{~g}^{-1} \mathrm{~K}^{-1}$ .
The glass of the thermometer and the beaker containing the water can be considered to have negligible heat capacity.

[ 4 ]

(i)

Calculate, to three significant figures, the final steady temperature indicated by the thermometer in the water. ${ }^{\circ} \mathrm{C}$

[ 3 ]

$Q=m c \Delta T$

C1

evidence of realisation that Q lost by water =Q gained by mercury

C1

$18.7 \times 4.18 \times(37.4-T)=6.94 \times 0.140 \times(T-23.0)$

C1

$T=37.2^{\circ} \mathrm{C}$

A1

(ii)

Suggest one change that could be made to the design of the thermometer that would enable it to give a more accurate measurement of temperature.

[ 1 ]

(c)

Explain why the thermometer in Fig. 2.1 does not provide a direct measurement of thermodynamic temperature.

[ 2 ]

depends on properties of a real substance

B1

$0^{\circ} \mathrm{C}$ is not absolute zero

B1

[Maximum number: 3]

A fixed mass of an ideal gas at a temperature of $20^{\circ} \mathrm{C}$ is sealed in a cylinder by a piston, as shown in Fig. 2.1.

Fig. 2.1

The initial volume of the gas is $1.24 \times 10^{-4} \mathrm{~m}^{3}$ .
Thermal energy is supplied to the gas and its volume increases by $5.20 \times 10^{-5} \mathrm{~m}^{3}$ .

The mass of the gas is 16 g . For this expansion, there is a net transfer of 960 J of thermal energy to the gas.

Calculate the specific heat capacity c of the gas at this pressure.

c=\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}

$c=Q / m \Delta T$

C1

$\begin{aligned} =960 /(0.016 \times(416-293)) =490 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1} \end{aligned}$

A1

All question bank results loaded

Covered syllabus points

14. Temperature question 1

Question (a)

Question (i)

Question (ii)

Question (b)

Question (c)

Question (i)

Question (ii)

14. Temperature question 2

Question (a)

Question (b)

Question (i)

Question (ii)

Question (c)

14. Temperature question 3