Q. 1 g of steam at 100$℃$ and equal mass of ice at 0$℃$ are mixed. The temperature of the mixture in steady state will be (latent heat of steam=540${calg}^{- 1}$, latent heat of ice=80${calg}^{- 1}$ )
A) 50$℃$
B) 100$℃$
C) 67$℃$
D) 33$℃$
Explanation
(b) Heat taken by ice to raise its temperature to 100$℃$
$\mathcal{Q}_{1} = 1 \times 80 + 1 \times 1 \times 100 = 180$ cal
Heat given by steam when condensed
$\mathcal{Q}_{2} = m_{2}L_{2} = 1 \times 540 = 540$ cal
As $\mathcal{Q}_{2} > \mathcal{Q}_{1}$, hence, temperature of mixture will remain 100$℃$.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.