Question

An Insulated vessel contains $0.4kg$ of water at $0^{o}C$. A piece of $0.1kg$ ice at $-{15}^{o}C$ is put into it and steam at ${100}^{o}C$ is bubbled into it until all ice is melted and finally the contents are liquid water at ${40}^{o}C$. Assume that the vessel does not give or take any heat and there is no loss of matter and heat to the surroundings. Specific heat of ice is $2.2\times {10}^{3}Jk{g}^{-1}{K}^{-1}$ heat of fusion of water is $333\times {10}^{3}Jk{g}^{-1}$ heat of vaporization of water is $2260\times {10}^{3}Jk{g}^{-1}$. The amount of steam that was bubbled into the water is :

• A. $34.7g$
• B. $236g$
• C. $0.023g$
• D. $48.01g$

Answer 1

The correct option is D $48.01g$

Given:

$S_i=2200$ $J/kg.K$

$S_w=4200$ $J/kg.K$

$L_f=3.33\times {10}^5$$J/kg$

$L_v=2.26\times {10}^6$$J/kg$

$m_i=0.1$ $kg$

$m_w=0.4$ $kg$

Let the mass of steam bubbled be $m$.

Heat released by steam to reach at ${40}^{o}C$,

$H_r=m{L}_v+m{S}_w(100-40)$

$\therefore$ $H_r=m(2.26\times {10}^6)+m\left(4200\right)\left(60\right)=2.512\times {10}^{6}J$

Heat absorbed:

$H_a=m_i\left(15\right)\left(S_i\right)+m_i{L}_f+(0.1+0.4)S_w\left(40\right)$

$\therefore$ $H_a=0.1\left(15\right)\left(2200\right)+0.1(3.3\times {10}^5)+\left(0.5\right)\left(4200\right)\left(40\right)=1.206\times {10}^{5}J$

As $H_a=H_r$

$\therefore$ $1.206\times {10}^5=m(2.512\times {10}^6)$

$⟹m=0.048$ $kg$ $=48.01$ $g$