Question

Steam at ${100}^{0}C$ is passed into $1.1kg$ of water contained in calorimeter of water equivalent $0.02kg$ at ${15}^{0}C$ till the temperature of the calorimeter rises to ${80}^{0}C$. What is the mass of steam condensed?
Latent heat of stem $=536cal/g$

• A. $0.31kg$
• B. $0.13kg$
• C. $0.41kg$
• D. $0.15kg$

Answer 1

The correct option is B $0.13kg$

The correct option is A

We have,

The mass of water and the calorimeter= $(1.10+0.02)=1.12kg$

The specific heat capacity of the water $=4.184\times {10}^{3}J/KgK$

Now, By using,

$Q=mc\Delta T$

Heat gained by the calorimeter and water is:

$Q=1.12\times 4.18\times {10}^3\times 65J/KgK.....1$

Let the mass of the steam be m kg,

So, latent heat of the vaporization of water at

${100}^{0}c=540cal/g=540\times 4.184\times {10}^{3}J/Kg$

Therefore,

Heat lost by the steam,

$Q^{\prime }=m(540\times 4.184\times {10}^3)+m(4.184\times {10}^3\times (100-80\left)\right)J/Kgk....2$

By equating the above equations then we get,

$m(540+20)=1.12\times 65$

$m=\frac{1.12\times 65}{560}=0.13kg$