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Calculate the heat required to convert $3\; kg$ of ice at $-12\,^{\circ} C$ kept in a calorimeter to steam at $100\,^{\circ} C$ at atmospheric pressure. Given specific heat capacity of $ice =2100\; J \,kg ^{-1} K ^{-1}$. specific heat capacity of water $=4186\; J kg ^{-1} K ^{-1}$, latent heat of fusion of ice $=3.35 \times 10^{5} \;J \,kg ^{-1}$ and latent heat of steam $=2.256 \times 10^{6}\; J \,kg ^{-1}$
$9.1 \times 10^{6} \;J$
$2.4 \times 10^{4} \;J$
$6.2 \times 10^{5} \;J$
$8.6 \times 10^{7} \;J$
Solution
Mass of the ice $ , m=3 kg$
specific heat capacity of ice $ , s _{\text {tee }}$
$=2100 J kg ^{-1} K ^{-1}$
specific heat capacity of water, $s_{\text {water }}$
$=4186 J kg ^{-1} K ^{-1}$
latent heat of fusion of ice $ , L_{ ice }$
$=3.35 \times 10^{5} J kg ^{-1}$
latent heat of steam, $L_{\text {steam }}$
$=2.256 \times 10^{6} J kg ^{-1}$
– Jow, $\quad Q=$ heat required to convert $3 kg$ of
$\text { 1ce at }-12^{\circ} C \text { to steam at } 100^{\circ} C$
$Q_{1}=$ heat required to convert ice at
$-12^{\circ} C \text { to } 1 ce \text { at } 0^{\circ} C$
$= m s _{\text {sce }} \Delta T_{1}=(3 kg )\left(2100 J kg ^{-1}\right.\left. K ^{-1}\right)[0-(-12)]^{\circ} C =75600 J$
$Q_{2}=$ heat required to melt $1 ce$ at $0^{\circ} C$ to water at $0^{\circ} C$
$=m L_{ fice }=(3 kg )\left(3.35 \times 10^{5} J kg ^{-1}\right)$
$=1005000 J$
$Q_{3}=$ heat required to convert water at $0^{\circ} C$ to water at $100^{\circ} C$
$=m s_{w} \Delta T_{2}=(3 kg )\left(4186 J kg ^{-1} K ^{-1}\right)$
$-\left(100^{\circ} C \right)$
$=1255800 \;J$
$Q _{4}=$ heat required to convert water at $100^{\circ} C$ to steam at $100^{\circ} C$
$=m L_{\text {steam }}=(3 kg )\left(2.256 \times 10^{6}\right.\left.J kg ^{-1}\right)$
$= 6768000 J$
$Q =Q_{1}+Q_{2}+Q_{3}+Q_{4}$
$=75600 J +1005000 J+1255800 J +6768000 J$
$=9.1 \times 10^{6} \;J$
