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(C) Let $m$ be the mass of water. The heat required to raise the temperature from $0 	ext{ °C}$ to $100 	ext{ °C}$ is $Q_1 = m \cdot c \cdot \Delta

Vedclass · Accepted Answer

$550$

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(C) Let $m$ be the mass of water. The heat required to raise the temperature from $0 	ext{ °C}$ to $100 	ext{ °C}$ is $Q_1 = m \cdot c \cdot \Delta T = m \cdot 1 \cdot (100 - 0) = 100m 	ext{ cal}$.Given that this process takes $10 	ext{ minutes}$,the rate of heat supply is $P = \frac{100m}{10} = 10m 	ext{ cal/min}$.Now,the water is converted into steam at $100 	ext{ °C}$ in $55 	ext{ minutes}$. The heat required for this phase change is $Q_2 = m \cdot L$.The heat supplied in $55 	ext{ minutes}$ is $Q_2 = P 	imes 55 = 10m 	imes 55 = 550m 	ext{ cal}$.Equating $Q_2 = mL$,we get $mL = 550m$,which implies $L = 550 	ext{ cal/g}$.

During an experiment,it takes $10 \text{ minutes}$ to raise the temperature of water from $0 \text{ °C}$ to $100 \text{ °C}$,and then it takes another $55 \text{ minutes}$ for the water to completely convert into steam. The value of the latent heat of vaporization is ....... $\text{cal/g}$.

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