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WAEC Physics 2020 Paper

Question : 34

Total: 51

Solution:

Concept:

Specific latent heat of vaporization is the heat required to change

1\, \mathrm{kg}

of a liquid into vapour at constant temperature.

Explanation:

From the heating curve, vaporization begins when the heat absorbed is

8 \times 10^{3}\, \mathrm{J}

Vaporization ends when the heat absorbed is

53 \times 10^{3}\, \mathrm{J}

So, the heat used only for vaporization is:

Q = 53 \times 10^{3} - 8 \times 10^{3} = 45 \times 10^{3}\, \mathrm{J}

The formula for latent heat is:

Q = mL

Here,

m = 0.02\, \mathrm{kg}

Therefore,

L = \frac{Q}{m} = \frac{45 \times 10^{3}}{0.02}

L = 2.25 \times 10^{6}\, \mathrm{J\,kg^{-1}}

Answer:

2.25 \times 10^{6}\, \mathrm{J\,kg^{-1}}

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