Question

When $0.15\,{\text{kg}}$at ${{\text{0}}^ \circ }{\text{C}}$mixed with of $0.30\,{\text{kg}}$water at ${50^ \circ }{\text{C}}$in a container, the resulting temperature is${6.7^ \circ }{\text{C}}$. Calculate heat of fusion of ice. $\left( {{{\text{S}}_{{\text{water}}}}\,{\text{ = }}\,{\text{4186}}\,{\text{J}}{\text{.k}}{{\text{g}}^{ - 1}}{{\text{K}}^{ - 1}}} \right)$

Answer 1

Heat of fusion is the heat required for the melting of ice heat which will be given by water. The heat lost by water will be equal to the heat gained by the ice.

Step by step answer: The heat of fusion represents the heat taken by a solid to convert into liquid. The heat which will be taken by solid, here, ice will be provided by some source, here, water. So, the solid will gain heat and the water will lose the heat.
So, lost heat by water will be equal to gained heat by ice.
The formula used to determine the heat change is as follows:
$q\, = \,mS\Delta T$
Where,
$q\,$ is the heat.
$m$ is the mass of the substance
$S$ is the heat capacity of the substance
$\Delta T$ is the change in temperature
Determine the heat loses by the water as follow:
Convert the temperature in kelvin by adding $273.15$ and then calculate the temperature change by subtracting the initial temperature from the final temperature.
$\Rightarrow \Delta {T_{water}}\, = \left( {273.15\, + 50} \right) - \left( {273.15\, + 6.7} \right)$
$\Rightarrow \Delta {T_{water}}\, = 43.3\,{\text{K}}\,$
Substitute $0.30\,{\text{kg}}$for the mass of water, ${\text{4186}}\,{\text{J}}{\text{.k}}{{\text{g}}^{ - 1}}{{\text{K}}^{ - 1}}$ for the heat capacity of water and $43.3\,{\text{K}}$ for temperature change.
${q_{water}}\, = \,0.30\,{\text{kg}} \times {\text{4186}}\,{\text{J}}{\text{.k}}{{\text{g}}^{ - 1}}{{\text{K}}^{ - 1}} \times 43.3\,{\text{K}}$
${q_{water}}\, = \,54376.14\,{\text{J}}$

The heat of fusion of ice will be the sum of latent heat and heat required for temperature change.
${q_{ice}}\, = \,mL\, + \,mS\Delta T$
Where,
$L$ is the latent heat.
$\Rightarrow \Delta {T_{ice}}\, = \left( {273.15\, + 6.7} \right) - \left( {273.15\, + 0} \right)$
$\Rightarrow \Delta {T_{ice}}\, = 6.7\,{\text{K}}\,$
Substitute $0.15\,{\text{kg}}$for the mass of ice, ${\text{4186}}\,{\text{J}}{\text{.k}}{{\text{g}}^{ - 1}}{{\text{K}}^{ - 1}}$ for the heat capacity of water and $6.7\,{\text{K}}$ for temperature change.
${q_{ice}}\, = \,0.15\,{\text{kg}} \times L\, + \,0.15\,{\text{kg}} \times {\text{4186}}\,{\text{J}}{\text{.k}}{{\text{g}}^{ - 1}}{{\text{K}}^{ - 1}} \times 6.7\,{\text{K}}$
${q_{ice}}\, = \,0.15\,{\text{kg}} \times L\, + 4206.93\,{\text{J}}$
Now, ${q_{water}}\, = \,{q_{ice}}$

So, $\,54376.14\,{\text{J}}\,\, = \,\,0.15\,{\text{kg}} \times L\, + 4206.93\,{\text{J}}$
$\Rightarrow$ $\,0.15\,{\text{kg}} \times L = 54376.14\,{\text{J}}\,\, - 4206.93\,{\text{J}}$
$\Rightarrow$ $\,0.15\,{\text{kg}} \times L = 50169.21\,{\text{J}}$
$\Rightarrow$ $\,L = \dfrac{{50169.21\,{\text{J}}}}{{0.15\,{\text{kg}}}}$
$\Rightarrow L = 334461.4\,{\text{J}}{\text{.K}}{{\text{g}}^{ - 1}}$

So, the heat of fusion is $3.3\, \times {10^5}\,{\text{J}}{\text{.K}}{{\text{g}}^{ - 1}}$.

Note: When ice converts into the liquid, two changes take place one is phase change and the second is temperature change. The heat used in phase change is known as latent heat. So, the heat of fusion of ice is the sum of latent heat and heat required for temperature change.