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An electric heater with constant heat supply rate is used to convert a certain amount of liquid ammonia to saturated vapour at high pressure. The heater takes 14 minutes to bring the liquid at $15^{\circ} \mathrm{C}$ to the boiling point of $50{ }^{\circ} \mathrm{C}$ and 92 minutes to convert the liquid at the boiling point wholly to vapour. If the specific heat capacity of liquid ammonia is $4.9 \mathrm{~kJ} / \mathrm{kg} \mathrm{K}$, the latent heat of vaporisation of ammonia in $\mathrm{kJ} / \mathrm{kg}$ is
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The correct answer is:
1127
Let mass of liquid ammonia taken is $m$ and rate of heat supplied is $r$.
Then, as per given condition, $r \times 14=m \times 4.9 \times(50-15)$ and $r \times 92=m L$
Dividing these, we get
$$
\begin{aligned}
\frac{92}{14} & =\frac{L}{4.9 \times(50-15)} \text { or } L=\frac{92 \times 4.9 \times 35}{14} \\
& =1127 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$
Then, as per given condition, $r \times 14=m \times 4.9 \times(50-15)$ and $r \times 92=m L$
Dividing these, we get
$$
\begin{aligned}
\frac{92}{14} & =\frac{L}{4.9 \times(50-15)} \text { or } L=\frac{92 \times 4.9 \times 35}{14} \\
& =1127 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$
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