WAEC - Physics (2020 - No. 34)

The sketched graph above illustrates the heating curve of a 0.02kg of water. Determine the approximate value of the specific latent heat of vapourization of water
 

2.25 x 10\(^6\) J kg\(^{-1}\)
4.17 x 10\(^3\) J kg\(^{-1}\)
2.00 x 10\(^3\) J kg\(^{-1}\)
1.00 x 10\(^3\) J kg\(^{-1}\)

Explanation

specific latent heat of vapourization is the amount of heat energy needed to change 1kg of liquid to vapour without a temperature change. The heat absorbed by the liquid at the beginning of vapourization is \(8 \times 10^3\)J and heat absorbed after the change of stage is \(53 \times 10^3\)J. therefore, the total heat energy needed to change the liquid completely to vapour is Q = \(53 \times 10^3 - 8 \times 10^3 J = 45 \times10^3\)J
Recall, Q = mL
Where, L = specific latent heat of vapourization

Q = mL

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

L = \(2.25 \times10^6JKg^{-1}\)

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