Question

$1g$ of steam at ${100}^{\circ }C$ and an equal mass of ice at $0^{\circ }C$ are mixed. The temperature of the mixture in steady state will be : (latent heat of steam $=540cal/g$, latent heat of ice $=80cal/g$, specific heat of water $=1cal/g^{\circ }C$)

• A. ${50}^{\circ }C$
• B. ${100}^{\circ }C$
• C. ${67}^{\circ }C$
• D. None of these

Answer 1

The correct option is B ${100}^{\circ }C$

Let the mixture become water at $T^{o}C$ at steady state.
To convert steam at ${100}^{o}C$ to water at $T^{o}C$
$Q=1\times 540+1\times 1\times (100-T)$

Similarly, to convert 1 g of ice to water at T C
$Q=1\times 80+1\times 1\times (T-0)$

Since, both these heats should be the same,
$540+100-T=80+T$
$⟹2T=560$
$⟹T={280}^{o}C$
Since, this is not a possible value of T hence water wont exist and it will be precisely steam at ${100}^{o}C$