The specific heat of water $=4200 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$ and the latent heat of ice $=3.4 \times 10^{5} \mathrm{~J} \mathrm{~kg}^{-1}$. 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
Here the water will provide heat for ice to melt therefore
$\mathrm{m}_{\mathrm{w}} \mathrm{s}_{\mathrm{w}} \Delta \theta=\mathrm{m}_{\mathrm{ice}} \mathrm{L}_{\mathrm{ice}}$
$\mathrm{m}_{\mathrm{ice}}=\frac{0.2 \times 4200 \times 25}{3.4 \times 10^{5}}$
$=0.0617 \mathrm{~kg}$
$=61.7 \mathrm{gm}$
Remaining ice will remain un-melted
so correct answer is 1
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