The specific heat of water $=4200 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$ and the latent heat of ice $=3.4 \times 10^{5} \mathrm{Jkg}^{-1} \cdot 100$ grams of ice at $0^{\circ} \mathrm{C}$ is placed in $200 \mathrm{~g}$ of water at $25^{\circ} \mathrm{C}$. The amount of ice that will melt as the temperature of water reaches $0^{\circ} \mathrm{C}$ is close to (in grams):
Correct Option: 1
(1) Here ice melts due to water.
Let the amount of ice melts $=m_{\text {ice }}$
$m_{w} s_{w} \Delta \theta=m_{\text {ice }} L_{\text {ice }}$
$\therefore m_{\text {ice }}=\frac{m_{w} s_{w} \Delta \theta}{L_{\text {ice }}}$
$=\frac{0.2 \times 4200 \times 25}{3.4 \times 10^{5}}=0.0617 \mathrm{~kg}=61.7 \mathrm{~g}$
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