Question

A thermally insulated vessel contains $150g$ of water at $0^{\circ }C$. Then the air from the vessel is pumped out adiabatically. A fraction of water turns into ice and the rest evaporates at $0^{\circ }C$ itself. The mass of evaporated water will be closest to (Latent heat of vaporization of water $=2.1\times {10}^{6}J/kg$ and Latent heat of Fusion of water $=3.36\times {10}^{5}J/kg$

• A. $35g$
• B. $150g$
• C. $130g$
  
• D. $20g$

Answer 1

The correct option is C $130g$


We know that: $Q=mL$

Given: $m_w=150g,L_v=2.1\times {10}^{6}J/kg$,
$L_f=3.36\times {10}^{5}J/kg$

Suppose '$m$' gram of water evaporates then, Mass that converts into ice is $(150–m)$

$\Delta Q_{released}=\Delta Q_{absorbed}$

$(150-m)L_f=m{L}_v$

$m(L_f+L_v)=150L_f$

$m=\frac{150L_f}{L_f+L_v}$

$m=\frac{150\times 3.36\times {10}^5}{3.36\times {10}^5+2.1\times {10}^6}$

$m=20g$

Final Answer: (a)