Question

Steam at ${100}^{\circ }C$ is passed into $1.1kg$ of water contained in calorimeter of water equivalent $0.02kg$ at ${15}^{\circ }C$ till the temperature of the calorimeter rises to ${80}^{0}C$. What is the mass of steam condensed?

• (A) 0.130
• (B) 0.065
• (C) 0.260
• (D) 0.135

Answer 1

The Correct Option is A

Explanation:
The amount of heat lost by the steam to condense into the water at $100°C$ and in raising the temperature of water and calorimeter to $80°C$ is $H_1=mL+mc\mathrm{\Delta }T$wherem: Mass of steam$L=540cal/g$ Latent heat of vaporization of steam$\mathrm{\Delta }T={100}^{\circ }C-{80}^{\circ }C={20}^{\circ }C$ Decrease in temperature$c=1$ cal $/g^{\circ }C$ Specific heat of waterNow, the amount of heat gained by water to increase its temperature from ${15}^{\circ }C$ to ${80}^{\circ }C$ is$H_2=1.1kg\times c\times (80-15{)}^{\circ }C$$\Rightarrow H_2=1.1\times {10}^{3}g\times 1cal/g^{\circ }C\times {65}^{\circ }C$$\Rightarrow H_2=71500cal$Again, the amount of heat gained by the calorimeter to increase its temperature from ${15}^{\circ }C$ to ${80}^{\circ }C$ is$H_3=0.02kg\times c\times (80-15{)}^{\circ }C$$\Rightarrow H_3=0.02\times {10}^{3}g\times 1cal/g^{\circ }C\times {65}^{\circ }C$$\Rightarrow H_3=1300cal$From the principle of calorimetry, we have$H_1=H_2+H_3$$\Rightarrow mL+(mc\times {20}^{\circ }C)=71500cal+1300cal$$\Rightarrow 540cal/g\times m+(m\times 1cal/g^{\circ }C\times {20}^{\circ }C)=72800cal$$\Rightarrow 560$ cal $/g\times m=72800cal$$\Rightarrow m=\frac{72800}{560}g=130g$$\Rightarrow m=0.130kg$Therefore, the mass of steam condensed is $0.130kg$.Hence, the correct option is (A).