Question

$0.3kg$ of steam at ${100}^{0}C$ is mixed with $6kg$ of water. After the steam condenses the temperature of the mixture rises to ${40}^{0}C$. The initial temperature of the water is:

• A. ${70}^{0}C$
• B. ${30}^{0}C$
• C. ${40}^{0}C$
• D. ${10}^{0}C$

Answer 1

Answer: (D)

Heat released by steam on condesation is $Q_1=0.3\times 2260\times {10}^{3}J$ ;
where, Latent heat of condensation of steam is : $2260\times {10}^{3}J/k{g}^{o}C$

Let the initial temperature of water be $T$.

Balancing the energy released and absorbed,

$Q_1+0.3\times 4200\times (100-40)=6\times 4200\times (40-T)$ ;
where spceific heat of water is: $4200J/k{g}^{o}C$

So, $40-T\approx 30\Rightarrow T={10}^{o}C$