Question

\[10g\] of ice at \[{{0}^{\circ }}C\] absorbs \[5460J\] of heat energy to melt and change to water at \[{{50}^{\circ }}C\]. Calculate the specific latent heat of fusion of ice. Specific heat capacity of water is \[4200JK{{g}^{-1}}{{K}^{-1}}\]
A. \[2168J{{g}^{-1}}\]
B. \[336J{{g}^{-1}}\]
C. \[43.36J{{g}^{-1}}\]
D. \[21.68J{{g}^{-1}}\]

Answer 1

Hint: Use the concept of calorimetry.
Use the formula of the specific heat.
Use the concept of energy conservation.

Complete step by step solution:
According to the law of calorimeters there is no heat loss to the surrounding. The heat transfer occurs between the components of the isolated system. Heat flows from higher temperature to the lower temperature.
Mass of the ice$=10g=0.01kg$
As we know that $1g=0.001kg$
Amount of the heat absorbed by the ice$=5460J$
Specific heat capacity of the water$=4200J/kgK$
We have to find the specific latent heat of fusion of the ice.
As we know that amount of heat needed/lost during a temperature change of $\Delta T$ for the mass $m$ having specific heat capacity as $c$ is given by,
$Q=mc\Delta T$
Amount of heat needed to raise the temperature of water from $0{}^\circ C\text{ to }50{}^\circ C$
$\begin{align}
  & {{Q}_{1}}=\left( 0.01 \right)\left( 4200 \right)\left( 50 \right)J \\
 & =2100J
\end{align}$
Let the specific latent heat of fusion of ice is $LJ/kg$
Then,
Total heat absorbed$=$ heat needed to melt the ice$+$heat needed to raise the temperature
$\begin{align}
  & Q=mL+mc\Delta T \\
 & 5460J=10L+2100J \\
 & 10L=3360J \\
 & L=\dfrac{3360}{10}J/g \\
 & =336J/g
\end{align}$
Hence, the specific latent heat of fusion of ice is $336J/g$

Note: As per law of calorimetry the heat energy gained is equal to the heat energy lost by the components of the calorimeter.
Temperature remains constant during change of the state.
Students should have an idea about the latent heat of fusion.