Question

The amount of heat energy required to convert 1 kg of ice
at $-{10}^{\circ }\text{C}$ completely into water at ${100}^{\circ }\text{C}is777000J$. Calculate the specific latent heat of ice. Specific heat capacity of ice is $2100Jk{g}^{-1}{K}^{-1}$, Specific heat capacity of water is $4200Jk{g}^{-1}K-1$ .

Answer 1

Step 1: Calculation of amount of heat gained by ice to raise its temperature to$0^{\circ }\text{C}$
Given,
Mass of ice, $m=1kg$
Initial temperature, $T_1=-{10}^{\circ }\text{C}=263K$
Final temperature, $T_2=0^{\circ }\text{C}=273K$
Specific heat capacity of ice, $c=2.1J{g}^{-1}{K}^{-1}$
Let heat energy $=Q_1$
We know,
Specific heat capacity,
$c=\frac{Q}{m\times \Delta T}$
$\Rightarrow Q=mc\Delta T$
$\Delta T=T_2-T_1$
$\Delta T=273K-263K=10K$
Therefore,
$\Rightarrow Q_1=\left(1kg\right)\times \left(2100Jk{g}^{-1}K-1\right)\left(10K\right)$
$\Rightarrow Q_1=21000J$

Step 2: Calculation of amount of heat required to convert ice into water at$0^{\circ }\text{C}$
Specific latent heat of ice, $L$
Specific latent heat,
$L=\frac{Q}{m}(atT=0^{\circ }\text{C})$
$\Rightarrow Q=mL$
Therefore,
$Q_2=\left(1kg\right)\times \left(L\right)$
$Q_2=L$

Step 3: Calculation of amount of heat gained by water to raise its temperature to ${100}^{\circ }\text{C}$
Initial temperature, $T_1=0^{\circ }\text{C}=273K$
Final temperature,$T_2={100}^{\circ }\text{C}=373K$
Specific heat capacity of water,$c=4200Jk{g}^{-1}{K}^{-1}$
$\Delta T=T_2-T_1$
$\Delta T=373K-273K=100K$
Therefore,
$Q=mc\Delta T$
$\Rightarrow Q_3=\left(1kg\right)\times \left(4200Jk{g}^{-1}{K}^{-1}\right)\times \left(100K\right)$
$\Rightarrow Q_3=420000J$

Step 4: Calculation of specific latent heat of ice
Total amount of heat energy,$Q_T=777000J$
$Q_T=Q_1+Q_2+Q_3$
$777000J=\left(21000J\right)+\left(L\right)+\left(420000J\right)$
$\Rightarrow L=777000-21000-420000$
$\Rightarrow L=336000JK{g}^{-1}$
The specific latent heat of ice is $336000JK{g}^{-1}$ .