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 water in the container at thermal equilibrium is____.Latent heat of vaporisation = 540 cal/gm. Specific heat of water = 1 cal/gm-${}^{0}C$.

• A. $18gm$
• B. $20gm$
• C. $22gm$
• D. $24gm$

Answer 1

The correct option is C $22gm$

Heat released by steam in conversion to water at ${100}^{0}C$ is
$Q_1$ = mL = 4 $\times$540 = 2160 cal.
Heat required to raise temperature of water from ${46}^{0}C$ to ${100}^{0}C$ is
$Q_2=mS\Delta \theta$ = 20 1 54 = 1080 J
$Q_1$ > $Q_2$ and $\frac{Q_1}{Q_2}$ = 2
Hence all steam is not converted to water only half steam shall be converted to water
Therefore, Final mass of water $=20+2=22$ gm