Question

Calculate the total amount of heat energy required to convert $100g$ of ice at$-{10}^{\circ }\text{C}$ completely into water at ${100}^{\circ }\text{C}$ . Specific heat capacity of ice is $2.1J{g}^{-1}{K}^{-1}$, specific heat capacity of water is $4.2J{g}^{-1}{K}^{-1}$, specific latent heat of fusion of ice is $336J{g}^{-1}$.

Answer 1

Step 1: Calculate the amount of heat gained by ice to raise its temperature to $0^{\circ }\text{C}$
Given,
Mass of ice,$m=100g$
Initial temperature, $T_1=-{10}^{\circ }\text{C}=263K$
Final temperature, $T_2=0^{\circ }\text{C}=273K$
Specific heat capacity of ice, $c=2.1J{g}^{-1}{K}^{-1}$
$\Delta T=T_2-T_1$
$\Delta T=273K-263K=10K$
We know, Specific heat capacity,
$c=\frac{Q}{m\times \Delta T}$
Therefore,
$Q=mc\Delta T$
$\Rightarrow Q_3=\left(100g\right)\times \left(2.1J{g}^{-1}{K}^{-1}\right)\times \left(10K\right)$
$\Rightarrow Q_1=2100J$

Step 2: Calculate the amount of heat required to convert ice into water at$0^{\circ }\text{C}$
Specific latent heat of fusion of ice$=336J{g}^{-1}$
Specific latent heat,
$L=\frac{Q}{m}(atT=0^{\circ }\text{C})$
$\Rightarrow Q=mL$
Therefore,
$Q_2=\left(100g\right)\times \left(336J{g}^{-1}\right)$
$Q_2=33600J$

Step 3: Calculate the 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,
$c=4.2J{g}^{-1}K-1$
$\Delta T=T_2-T_1$
$\Delta T=373K-273K=100K$
Therefore,
$Q=mc\Delta T$
$\Rightarrow Q_3=\left(100g\right)\times \left(4.2J{g}^{-1}{K}^{-1}\right)\times \left(100K\right)$
$\Rightarrow Q_3=42000J$

Step 4: Calculate total amount of heat energy
Total amount of heat energy,$Q=Q_1+Q_2+Q_3$
$Q=\left(2100J\right)+\left(33600J\right)+\left(42000J\right)$
$Q=77700J$
$Q=7.77\times {10}^{4}J$
Total amount of heat energy required is $7.77\times {10}^{4}J$.