Question

Steam at ${100}^{0}C$ is passed into $20grams$ of water at ${10}^{0}C$. When resultant temperature is ${80}^{0}C$, then weight of the water present is

• A. $27.33g$
• B. $24.8g$
• C. $22.5g$
• D. $30g$

Answer 1

The correct option is C $22.5g$

heat lost by the steam = heat gained by water

$m\times L=ms\Delta t$

$m\times 540=20\times 1\times (80-10)$

$\therefore m=\frac{20\times 70}{540}=2.5g$

Therefore the net amount of wate $=20+2.5=22.5g$