Question

A molten metal of mass $150\,{\text{g}}$ is kept at its melting point $800^\circ {\text{C}}$. When it is allowed to freeze at the same temperature, it gives out of $75000\,{\text{J}}$ heat energy.
(a) What is the specific latent heat of the metal?
(b) If the specific heat capacity of metal is $200\,{\text{J}} \cdot {\text{k}}{{\text{g}}^{ - 1}} \cdot {{\text{K}}^{ - 1}}$, how much additional heat energy will the metal give out in cooling to $- 50^\circ {\text{C}}$?

Answer 1

Use the formula for the latent heat given out in freezing in terms of the specific latent heat. Also use the formula for the energy or heat exchanged in terms of mass of the substance, specific heat of the substance and change in temperature of the substance. Using the formula for latent heat, calculate specific latents of the metal. Then use the second formula and calculate the amount of heat energy given out.

Formulae used:
The latent heat $Q$ of fusion is given by
$Q = mL$ …… (1)
Here, $m$ is the mass of the substance and $L$ is the latent heat of the substance.
The heat energy $Q$ exchanged by the substance is given by
$Q = mc\Delta T$ …… (2)
Here, $m$ is the mass of the substance, $c$ is specific heat of the substance and $\Delta T$ is a change in temperature of the substance.

Complete step by step answer:
We have given that the mass of the molten metal is $150\,{\text{g}}$ and melting temperature of the metal is $800^\circ {\text{C}}$.
$m = 150\,{\text{g}}$
$\Rightarrow T = 800^\circ {\text{C}}$
When the molten metal freezes $75000\,{\text{J}}$ of heat energy is given out.
${Q_f} = 75000\,{\text{J}}$

(a) We have asked to calculate the specific latent heat of the metal.When the metal freezes and solidifies, the temperature of the metal decreases and the metal gives out the energy equal to the latent heat.We can calculate the latent heat of freezing using equation (1).Rewrite equation (1) for the latent heat of freezing of the metal.
${Q_f} = mL$
$\Rightarrow L = \dfrac{{{Q_f}}}{m}$
Substitute $75000\,{\text{J}}$ for ${Q_f}$ and $150\,{\text{g}}$ for $m$ in the above equation.
$\Rightarrow L = \dfrac{{75000\,{\text{J}}}}{{150\,{\text{g}}}}$
$\Rightarrow L = \dfrac{{75000\,{\text{J}}}}{{150 \times {{10}^{ - 3}}\,{\text{kg}}}}$
$\therefore L = 5 \times {10^5}{\text{J}} \cdot {\text{k}}{{\text{g}}^{ - 1}}$

Hence, the specific latent heat of metal is $5 \times {10^5}{\text{J}} \cdot {\text{k}}{{\text{g}}^{ - 1}}$.

(b) We have given the specific heat capacity of the metal as $200\,{\text{J}} \cdot {\text{k}}{{\text{g}}^{ - 1}} \cdot {{\text{K}}^{ - 1}}$.
$c = 200\,{\text{J}} \cdot {\text{k}}{{\text{g}}^{ - 1}} \cdot {{\text{K}}^{ - 1}}$
We have asked to calculate the energy given out by the metal I cooling to the temperature $- 50^\circ {\text{C}}$.
${T_f} = - 50^\circ {\text{C}}$
Let us first calculate the change in temperature of the metal.
$\Delta T = T - {T_f}$
$\Rightarrow \Delta T = \left( {800^\circ {\text{C}}} \right) - \left( { - 50^\circ {\text{C}}} \right)$
$\Rightarrow \Delta T = 850^\circ {\text{C}}$
Substitute $150\,{\text{g}}$ for $m$, $200\,{\text{J}} \cdot {\text{k}}{{\text{g}}^{ - 1}} \cdot {{\text{K}}^{ - 1}}$ for $c$ and $850^\circ {\text{C}}$ for $\Delta T$ in equation (2).
$Q = \left( {150\,{\text{g}}} \right)\left( {200\,{\text{J}} \cdot {\text{k}}{{\text{g}}^{ - 1}} \cdot {{\text{K}}^{ - 1}}} \right)\left( {850^\circ {\text{C}}} \right)$
$\Rightarrow Q = \left( {0.150\,{\text{kg}}} \right)\left( {200\,{\text{J}} \cdot {\text{k}}{{\text{g}}^{ - 1}} \cdot {{\text{K}}^{ - 1}}} \right)\left( {850^\circ {\text{C}}} \right)$
$\therefore Q = 25500\,{\text{J}}$

Hence, the additional heat given out by the metal is $25500\,{\text{J}}$.

Note: The students may think that the specific heat of the metal is given in terms of unit kelvin for temperature then why we have used the temperature of the metal in degree Celsius. But the students can observe that when the unit of temperature of the metal is converted to degree kelvin then also the temperature difference remains the same and we will end with the same answer. That’s why we have not converted the unit of temperature of the metal.