Question

The graph shown in the figure represent change in the temperature of 5 kg of a substance as it abosrbs heat at a constant rate of 42 kJ (min$^{-1}$) The latent heat of vapourazation of the substance is :

• A. 630 kJ kg$^{-1}$
• B. 126 kJ kg$^{-1}$
• C. 84 kJ kg$^{-1}$
• D. 12.6 kJ kg$^{-1}$

Answer 1

Answer: (C)

84 kJ kg$^{-1}$

Answer 2

The graph shows a plateau from 20 to 30 minutes, representing vaporization at a constant temperature. The duration of this phase change is 10 minutes. At a heating rate of 42 kJ/min, the total heat absorbed during vaporization is $42 \text{ kJ/min} \times 10 \text{ min} = 420 \text{ kJ}$. For a mass of 5 kg, the latent heat of vaporization ($L_v$) is calculated using $Q = m L_v$, resulting in $L_v = \frac{420 \text{ kJ}}{5 \text{ kg}} = 84 \text{ kJ/kg}$.