100 g of ice at -10^circC is added to 200 g…

100 g of ice at -10^circC is added to 200 g of water at 40^circC in an insulated container of negligible heat capacity. Specific heat of ice = 0.5…

Single correct option

100\ \mathrm{g}

of ice at

-10^\circ\mathrm{C}

is added to

200\ \mathrm{g}

of water at

40^\circ\mathrm{C}

in an insulated container of negligible heat capacity. Specific heat of ice

= 0.5\ \mathrm{cal\,g^{-1}\,{}^\circ C^{-1}}

, specific heat of water

= 1\ \mathrm{cal\,g^{-1}\,{}^\circ C^{-1}}

, latent heat of fusion

= 80\ \mathrm{cal\,g^{-1}}

. When equilibrium is reached, the mass of ice that remains is

Show answer and solution

Answer: (A) $6.25\ \mathrm{g}$

Heat available from water cooling to

0^\circ\mathrm{C}

200\times1\times40 = 8000\ \mathrm{cal}

. Heat needed to warm the ice to

0^\circ\mathrm{C}

100\times0.5\times10 = 500\ \mathrm{cal}

. Melting all the ice would need a further

100\times80 = 8000\ \mathrm{cal}

, which exceeds the remaining

7500\ \mathrm{cal}

. So the final temperature is

0^\circ\mathrm{C}

with some ice left. Ice melted

= 7500/80 = 93.75\ \mathrm{g}

, so the ice remaining is

100 - 93.75 = 6.25\ \mathrm{g}

. A common error is to forget the

500\ \mathrm{cal}

for warming the ice and answer

0\ \mathrm{g}

12.5\ \mathrm{g}

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