Question

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

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

Answer 1

The correct option is C $22gm$

Given: 4 gms of steam at ${100}^{\circ }C$ is added to 20 gms of water at ${46}^{\circ }C$ in a container of negligible mass. Assuming no heat is lost to surrounding,

To find the mass of water in container at thermal equilibrium

Solution:

As per the given condition

Latent heat of vaporisation $L=540cal/gm.$

Specific heat of water $s=1cal/\left(g{m}^{\circ }C\right).$

Mass of the steam, $m_s=4g$

Mass of the water, $m_w=20g$

Heat present in steam $=m_{s}L=4\times 540cal=2160cal$

Heat required by water to raise temperature from ${46}^{\circ }C$ to ${100}^{o}C$ is

$=m_{w}s\left(\Delta T\right)⟹=20\times 1\times (100-46)=1080cal$

Heat present in steam is twice the heat required by water.

So, half of steam changes to water.

So new mass of water = $m_w+\frac{m_s}{2}=20+2=22gm$