Question

4 gms of steam at ${100}^{0}C$ is added to 20 gms of water at ${46}^{0}C$ in a container of negligible mass. Assuming no heat is lost to surrounding, the mass of steam (in gm) in container at thermal equilibrium is: (Latent heat of vaporisation $=$ 540 cal/gm. Specific heat of water $=1cal/gm$-${}^{0}C)$

• A. 2
• B. 3.42
• C. 4
• D. 0.82

Answer 1

Answer: (A)

2

Heat released by steam inconversion to water at ${100}^{\circ }C$ is
$Q_1=mL=4\times 540=2160cal$
Heat required to raise temperature of water from ${46}^{\circ }C$ to ${100}^{\circ }C$ is
$Q_2=mS\Delta \theta =20\times 1\times 54=1080cal$
Thus we see that $1080=2\left(540\right)$, i.e. only $2gm$ of steam is being used up in the heat transfer.

Thus the remaining mass of the steam is $4-2=2gm$