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Steam at $100\,^oC$ is passed into $1.1\,kg$ of water contained in calorimeter of water equivalent $0.02\ kg$ at $15\,^oC$ till the temperature of the calorimeter rises to $80\,^oC$. The mass of steam condensed in kilogram is
$0.13$
$0.065$
$0.260$
$0.135$
Solution
We have,
The mass of water and the calorimeter $=(1.10+0.02)=1.12\,kg$
The specific heat capacity of the water $=4.184 \times 10^3\,J / KgK$
Now, By using,
$Q=m c \Delta T$
Heat gained by the calorimeter and water is:
$Q =1.12 \times 4.18 \times 10^3 \times 65\,J / KgK \ldots . .1$
Let the mass of the steam be $m kg$,
So, latent heat of the vaporization of water at
$100^{\circ} C =540\,cal / g =540 \times 4.184 \times 10^3\,J / Kg$
Therefore,
Heat lost by the steam,
$Q^{\prime}=m\left(540 \times 4.184 \times 10^3\right)+m\left(4.184 \times 10^3 \times(100-80)\right) J / K g k \quad 2$
By equating the above equations then we get,
$m (540+20)=1.12 \times 65$
$m =\frac{1.12 \times 65}{560}=0.13\,kg$
