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Steam at $100°C$ is passed 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
$0.13$
$0.065$
$0.26$
$0.135$
Solution
(a) Heat is lost by steam in two stages
$(i)$ for change of state from steam at $100°C$ to water at $100°C$ is $m \times 540$
$(ii)$ to change water at $100°C $ to water at $80°C$ is $m \times 1 \times (100 -80),$ where $m$ is the mass of steam condensed.
Total heat lost by steam is $m \times 540 + m \times 20 = 560\, m (cals)$ Heat gained by calorimeter and its contents is
$= (1.1 + 0.02) \times (80 -15) = 1.12 \times 65 \,cals.$
using Principle of calorimetery, Heat gained $=$ heat lost
$560\, m = 1.12 \times 65, m = 0.130 \,gm$
