Question

In an insulated vessel, $0.05kg$ steam at $373K$ and $0.45kg$ of ice at $253K$ are mixed. Then, find the final temperature of the mixture.
Given, $L_{fusion}=80cal/g=336J/g$, $L_{vaporisation}=540cal/g=2268J/g$, $S_{ice}=2100J/kgK=0.5cal/gK$ and $S_{water}=4200J/kgK=1cal/gK$

• A. $273K$
• B. $373K$
• C. $100K$
• D. $0K$

Answer 1

The correct option is A $273K$

Heat absorbed in heating up of ice is:

$H_1=m_{ice}{S}_{ice}\Delta T$

$=0.05\times 2100\times 20=2100cal$

Heat absorbed in melting,

$H_2=m_{ice}{L}_{fusion}$

$=0.45\times 80\times {10}^3=36000cal$

Heat released in condensation of steam,

$H_3=m_{steam}{L}_{vaporisation}$

$=0.05\times 540\times {10}^3=27000cal$

Heat released in cooling water at $373K$ to water at $273K$ is:

$H_4=m_{steam}{S}_{water}\Delta T$

$=0.05\times 1000\times 100=5000cal$

Since, $H_4+H_3>H_2+H_1$, final temperature of mixture is $273K$