Question

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

• A. 20 gm of water at $0^{\circ }C.$
• B. 15 gm of water at $0^{\circ }C.$
• C. 10 gm ice and 10 gm of water at $0^{\circ }C.$
• D. 5 gm ice and 15 gm of water $0^{\circ }C.$

Answer 1

The correct option is C 10 gm ice and 10 gm of water at $0^{\circ }C.$

As we know $Q=mc\Delta \theta$

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

$Q_2=10\times 0.5(0-(-20\left)\right)+10\times 80=(100+800)cal=900cal$

As $Q_1<Q_2$,

so ice will not completely melt and final temperature = $0^{o}C$

As heat given by water in cooling up to $0^{o}C$ is only just sufficient to increase the temperature of ice from $-{20}^{o}C$ to $0^{o}C$, hence mixture in equilibrium will consist of 10gm of ice and 10gm of water, both at $0^{o}C$.