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$150 \mathrm{~g}$ of ice is mixed with $100 \mathrm{~g}$ of water at temperature $80^{\circ} \mathrm{C}$. The latent heat of ice is $80 \mathrm{cal} / \mathrm{g}$ and the specific heat of water is $1 \mathrm{cal} / \mathrm{g}-{ }^{\circ} \mathrm{C}$. Assuming no heat loss to the environment, the amount of ice which does not melt is -
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The correct answer is:
$50 \mathrm{~g}$
Heat loss by water = heat gain by ice.
$100 \times 1 \times 80=\mathrm{m} \times 80$
$\mathrm{m}=100 \mathrm{gm}$ ice melt
$\therefore$ Remaining ice $=50 \mathrm{~g}$
$100 \times 1 \times 80=\mathrm{m} \times 80$
$\mathrm{m}=100 \mathrm{gm}$ ice melt
$\therefore$ Remaining ice $=50 \mathrm{~g}$
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