(a) During melting, a solid becomes liquid with little or no change in volume. Use kinetic theory to explain

Q#4 (Past Exam Paper – June 2018 Paper 42 Q3)

(a) During melting, a solid becomes liquid with little or no change in volume.

Use kinetic theory to explain why, during the melting process, thermal energy is required

although there is no change in temperature. [3]

(b) An aluminium can of mass 160 g contains a mass of 330 g of warm water at a temperature of 38 °C, as illustrated in Fig. 3.1.

Fig. 3.1

A mass of 48 g of ice at –18 °C is taken from a freezer and put in to the water. The ice melts and the final temperature of the can and its contents is 23 °C.

Data for the specific heat capacity c of aluminium, ice and water are given in Fig. 3.2.

c / J g^-1 K^-1

aluminium       0.910

ice                   2.10

water               4.18

Fig. 3.2

Assuming no exchange of thermal energy with the surroundings,

(i) show that the loss in thermal energy of the can and the warm water is 2.3 × 10⁴ J, [2]

(ii) use the information in (i) to calculate a value L for the specific latent heat of fusion of ice. [2]

[Total: 7]

Solution:

(a) (During melting,) the bonds between atoms/molecules are broken, so the potential energy of atoms/molecules is increased. 

{Since the volume of the substance does not change – work done = pΔV = 0}

No/little work is done, so the required input of energy is thermal energy.

{The input of energy cannot be through work being done on the system.}

(b)

(i)

{Heat loss occurs both in the aluminium can and the warm water.

Heat loss of can = 160 × 0.910 × (38-23)

Heat loss of water = 330 × 4.18 × (38-23)}

(ΔQ =) mcΔθ

loss = (160 × 0.910 × 15) + (330 × 4.18 × 15)

loss = 2.3 × 10⁴ J

(ii)

{The heat loss by the can and warm water is gained by the ice to:

(1) increase its temperature (of ice) from -18°C to 0°C         [Q = 48 × 2.10 × (0- -18)]

(2) change its state from solid to liquid (melting)                    [Q = mL = 48 L]

(3) raise temperature of liquid from 0°C to 23°C                   [Q = 48 × 4.18 × (23-0)]}

2.3 × 10⁴ = (48 × 2.10 × 18) + 48L + (48 × 4.18 × 23)

48L = 1.66 × 10⁴

Specific latent heat of fusion, L = 350 J g^-1    

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