540, g of ice at 0,^oC is mixed with 540, g water at 80,^oC . final temperature of mixture is

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10-1.Thermometry, Thermal Expansion and Calorimetry

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$540\, g$ of ice at $0\,^oC$ is mixed with $540\, g$ water at $80\,^oC$ . The final temperature of the mixture is

A

$0\,^oC$

B

$40\,^oC$

C

$80\,^oC$

D

less than $0\,^oC$

Solution

The amount of heat required to melt $540 \mathrm{g}$ of ice is given by

$=540 \times 80=43200 \mathrm{cal}$

(latent heat of ice $=80 \mathrm{kcal} / \mathrm{g}$ )

Now heat lost by $540 \mathrm{g}$ of water form $80^{\circ} \mathrm{C}$ to $0^{\circ} \mathrm{C}$

$=540 \times 1 \times\left(80^{\circ}-0^{\circ}\right)$

$=43200 \mathrm{cal}$

Therefore, all the ice must be melt and temperature of mixture is zero.

Standard 11

Physics

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