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22320 cal of heat is supplied to 100 g of ice at $0^{\circ} \mathrm{C}$. If the latent heat of fusion of ice is 80 cal $\mathrm{g}^{-1}$ and latent heat of vaporization of water is 540 cal $\mathrm{g}^{-1},$ the final amount of water thus obtained and its temperature respectively
are
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are
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Verified Answer
The correct answer is:
$8 \mathrm{g}, 100^{\circ} \mathrm{C}$
Heat required to convert ice to water at $100^{\circ} \mathrm{C}$
$$
Q=m \times L+m s \Delta T=18000 \mathrm{cal}
$$
Amount of heat left $=4320 \mathrm{cal}$
$$
\begin{array}{l}
m \times L=4320 \\
m=8 g \text { steam }
\end{array}
$$
$$
Q=m \times L+m s \Delta T=18000 \mathrm{cal}
$$
Amount of heat left $=4320 \mathrm{cal}$
$$
\begin{array}{l}
m \times L=4320 \\
m=8 g \text { steam }
\end{array}
$$
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