# Question 2f390

Dec 25, 2016

$67.5 g$

#### Explanation:

We know that in such interactions heat is gained by one and heat is lost is by the other.

Also that the heat gained/lost is given by
$\Delta Q = m s t$, or $\Delta Q = m L$
where $m , s \mathmr{and} t$ are the mass, specific heat and rise or gain in temperature of the object;
$L$ is the latent heat for the change of state and
$\Delta {Q}_{\text{lost"=Delta Q_"gained}}$

In the given problem heat is lost by steam to condense and gained by ice which melts.

Suppose $m$ gram of ice at ${0}^{\circ} C$ melt as required.

When ice melts first it becomes water at ${0}^{\circ} C$. Similarly, when steam at ${100}^{\circ} C$ condenses it becomes water at same temperature. Thus we have a mixture of water at two different temperatures in the container. This is not practical. The mixture must have uniform temperature.

For sake of this problem, heat of melting of ice is being equated with heat of condensation of steam, i.e., change of phase in both cases. Latent heat for condensation of water is 540 calg^-1°C^-1 and the latent heat of fusion of ice is 80 calg^-1°C^-1#.

1. Heat released by condensation of steam
$\Delta Q = m {L}_{\text{condesation}} = 10.0 \times 540 = 5400 c a l$
2. Heat required to melt ice
$\Delta Q = m {L}_{\text{fusion}} = m \times 80 = 80 m c a l$
Equating both we get
$5400 = 80 m$
$\implies m = \frac{5400}{80} = 67.5 g$