Q: Steam at 100°C is passes into 1.1 kg of water contained in a calorimeter of water equivalent 0.02 kg at 15°C till the temperature of the calorimeter and its contents rises to 80°C. The mass of the steam condensed in kg is
Sol: Heat required
$\large Q_1 = (1.1+0.02)\times 10^3 \times 1 \times (80-15)$
Q1 = 72000 cal
Let mass of steam condensed is m kg , when heat is given .
Heat given $\large Q_2 = ( m \times 540 \times 10^3 ) + m \times 1 \times 10^3 \times (100-80)$
On applying , $\large Q_1 = Q_2 $
we get , m = 0.130 kg