Question

10 g of ice at $-{20}^{\circ }C$ is dropped into a calorimeter containing 10 g of water at ${10}^{\circ }C$; the specific heat of water is twice that of ice. Ehen equilibrium is reached, the calorimeter will contain

• A. 20 g of water
• B. 20 g of rice
• C. 10 g ice and 10 g of water
• D. 5 g ice and 15 g of water

Answer 1

The correct option is C

10 g ice and 10 g of water

$Q_1$ = $10\times 1\times 10$ = 100 cal

$Q_2$ = $10\times 0.50[0-(-20\left)\right]+10\times 80$

= ($100+800$) cal = $900$ cal

As $Q_1$ $<$ $Q_2$, so ice will not completely melt and final temperature = $0^{\circ }C$. As heat given by water in cooling up to $0^{\circ }C$ is only just sufficient to increase the temperature of the ice from $-{20}^{\circ }C$ to $0^{\circ }C$, hence mixture in equilibrium will consist of 10 g of ice and 10 g of water, both at $0^{\circ }C$.