Question

$60g$ of ice at $0^0$ is mixed with $60g$ of steam at ${100}^0$. At thermal equilibrium, the mixture contains:(Latent heats of steam and ice are $540cal{g}^{-1}$ and $80cal{g}^{-1}$ respectively, specific heat of water $=1cal{g}^{-1}{}^{-1}$)

• A. $80g$ of water and $40g$ of steam at ${100}^0$
• B. $120g$ of water at ${90}^0$
• C. $120g$ of water at ${100}^0$
• D. $40g$ of steam and $80g$ of water at $0^0$

Answer 1

The correct option is A $80g$ of water and $40g$ of steam at ${100}^0$

Heat energy required to melt $60g$ of ice $=60\times 80=4800Cal$.
Heat energy required to raise the temprature of water at $0$ degree centigrade to $100$ degree centigrade $=100\times 1\times 60=6000Cal$
Total energy required $=4800+6000=10800Cal$
This energy is given by the steam by getting converted into water.
Amount of steam getting converted into water is $\frac{10800}{540}=20g$
Thus, finally total $80gm$ of water will be there and $40gm$ of steam.