What is the amount of heat energy required to convert ice of mass 20kg at $-{{4}^{\circ }}C$ to water at ${{20}^{\circ }}C$? Use the relevant option for calculation.
a) Latent heat of fusion of ice $3.34\times {{10}^{5}}Jk{{g}^{-1}}$.
b) Specific heat capacity of water is $4180Jk{{g}^{-1}}{{K}^{-1}}$.
c) Specific heat of ice $2093Jk{{g}^{-1}}{{K}^{-1}}$

Hint: Use the formula for the heat required to raise the temperature of a body of mass m and specific heat capacity c by $\Delta T$ to calculate the heat energy required to raise the temperature of ice and water. Also use the formula for phase conversion of a given body.

Formula used:
$Q=mc\Delta T$
$Q=mL$

Complete step by step answer:
To convert a mass of ice at $-{{4}^{\circ }}C$ to water at ${{20}^{\circ }}C$, we must first provide some energy (${{Q}_{1}}$) to raise the temperature of the ice to ${{0}^{\circ }}C$. Then some heat energy (${{Q}_{2}}$) is needed to convert its phase from ice to water at ${{0}^{\circ }}C$. Again some energy (${{Q}_{3}}$) is needed to raise the temperature of water to ${{20}^{\circ }}C$.
The heat energy required to raise the temperature of a body of mass m by a temperature $\Delta T$is given as $Q=mc\Delta T$, where c is the specific heat of the body.
The heat energy required to convert a solid to liquid is given as $Q=mL$, where L is the latent heat of fusion.
Let's calculate the values of ${{Q}_{1}},{{Q}_{2}},{{Q}_{3}}$.
For calculating ${{Q}_{1}}$,
m = 20kg, $c=2093Jk{{g}^{-1}}{{K}^{-1}}$ and $\Delta T=4K$.
$\Rightarrow {{Q}_{1}}=20\times 2093\times 4=1.6744\times {{10}^{5}}J$
For calculating ${{Q}_{2}}$,
m = 20kg and $L=3.34\times {{10}^{5}}Jk{{g}^{-1}}$.
$\Rightarrow {{Q}_{2}}=20\times 3.34\times {{10}^{5}}=66.8\times {{10}^{5}}$
For calculating ${{Q}_{3}}$,
m = 20kg, $c=4180Jk{{g}^{-1}}{{K}^{-1}}$ and $\Delta T=20K$.
$\Rightarrow {{Q}_{3}}=20\times 4180\times 20=16.72\times {{10}^{5}}J$
Therefore, the total heat energy required to convert ice of mass 20kg at $-{{4}^{\circ }}C$ to water at ${{20}^{\circ }}C$ is ${{Q}_{1}}+{{Q}_{2}}+{{Q}_{3}}=1.6744\times {{10}^{5}}+66.8\times {{10}^{5}}+16.72\times {{10}^{5}}=85.1944\times {{10}^{5}}J$.

Note:
Note that when we want to raise the temperature of a body, we have to supply some amount of heat energy to the body. And when we want to lower the temperature of the body, we have to take some amount of heat energy from the body.
When these substances convert from solid to liquid or liquid to gas, we must give some heat.
When their substance converts from liquid to solid or gas to liquid, we must take some heat from the substance.