Question

In an experiment, a sphere of aluminium of mass $0 \cdot 2{\text{kg}}$ is heated up to $150^\circ {\text{C}}$ . Immediately it is then put into the water of volume $150{\text{cc}}$ at $27^\circ {\text{C}}$ kept in a calorimeter of water equivalent of $0 \cdot 025{\text{kg}}$ . The final temperature of the system is $40^\circ {\text{C}}$ . Find the specific heat of aluminium. (Take $4 \cdot 2{\text{Joule}} = 1{\text{calorie}}$ ).
A) $434{\text{Jk}}{{\text{g}}^{ - 1}}^\circ {{\text{C}}^{ - 1}}$
B) ${\text{378Jk}}{{\text{g}}^{ - 1}}^\circ {{\text{C}}^{ - 1}}$
C) $476{\text{Jk}}{{\text{g}}^{ - 1}}^\circ {{\text{C}}^{ - 1}}$
D) ${\text{315Jk}}{{\text{g}}^{ - 1}}^\circ {{\text{C}}^{ - 1}}$

Answer 1

Hint: The heated sphere is placed in a calorimeter of water to cool down. As a result, the heat of the sphere will be transferred to the calorimeter and the water in it. Thus the heat lost by the aluminium sphere will be the sum of the heat gained by the water and the container.

Formulas used:
-The amount of heat lost or heat gained by a sample is given by, $\Delta Q = mc\Delta T$ where, $m$ is the mass of the sample, $s$ is the specific heat of the sample and $\Delta T$ is the change in temperature of the sample.
-The mass of a substance is given by, $m = V \times \rho $ where $V$ is the volume of the substance and $\rho $ is the density of the substance.

Complete step by step solution:
Step 1: List the parameters known from the question.
The mass of the aluminium sphere that is heated is given to be ${m_a} = 0 \cdot 2{\text{kg}}$ and the temperature to which it is heated is ${T_a} = 150^\circ {\text{C}}$ . However, as it is placed in the water, its temperature decreases to $T = 40^\circ {\text{C}}$ .

The volume of water in which the sphere is kept is $V = 150{\text{cc}}$ and the initial temperature of the water is ${T_w} = 27^\circ {\text{C}}$ and finally, the temperature becomes $T = 40^\circ {\text{C}}$ .

The water equivalent of the calorimeter is given to be $w = 0 \cdot 025{\text{kg}}$ .

Also given $4.2{\text{Joule}} = 1{\text{calorie}}$ .

Step 2: Express the relation for the heat lost by the aluminium sphere.
The amount of heat lost by the aluminium sphere is given by, $\Delta {Q_{lost}} =
{m_a}{s_a}\Delta T$ ------ (1)
where ${m_a}$ is the mass of the sphere, ${s_a}$ is the specific heat of the aluminium and $\Delta T$ is the change in temperature of the sphere.

The change in temperature of the aluminium sphere is $\Delta T = {T_a} - T = 150 - 40 = 110^\circ {\text{C}}$

Substituting for ${m_a} = 0 \cdot 2{\text{kg}}$ and $\Delta T = 110^\circ {\text{C}}$ in equation (1) we get, $\Delta {Q_{lost}} = 0 \cdot 2 \times {s_a} \times 110$
Thus the amount of heat lost is $\Delta {Q_{lost}} = 22 \times {s_a}$ ------- (2)

Step 3: Express the relation for the heat gained.

The amount of heat gained is given by, $\Delta {Q_{gain}} = \left( {{m_w} + w} \right){s_w}\Delta T$ ------ (3)
where ${m_w} + w$ is the mass of the water, ${s_w}$ is the specific heat of the water and $\Delta T$ is the change in temperature of the calorimeter.

The change in temperature of the calorimeter is $\Delta T = {T_w} - T = 40 - 27 = 13^\circ {\text{C}}$ .

The mass of water in which the sphere is placed is given by, ${m_w} = V \times \rho $ where $V = 150 \times {10^{ - 6}}{{\text{m}}^3}$ is the volume of the water and $\rho = {10^3}{\text{kg}}{{\text{m}}^{ - 3}}$ is the density of water.
So the mass of water will be ${m_w} = 150 \times {10^{ - 6}} \times {10^3} = 150 \times
{10^{ - 3}}{\text{kg}}$

Substituting for ${m_w} = 150 \times {10^{ - 3}}{\text{kg}}$ , $w = 0 \cdot 025{\text{kg}}$ , ${s_w} = 4 \cdot 2 \times {10^3}{\text{Jk}}{{\text{g}}^{ - 1}}^\circ {{\text{C}}^{ - 1}}$and $\Delta T = 13^\circ {\text{C}}$ in equation (3) we get, $\Delta {Q_{gain}} = \left( {0 \cdot 15 + 0 \cdot 025} \right) \times 4 \cdot 2 \times {10^3} \times 13 = 9555{\text{J}}$

Thus the amount of heat lost is $\Delta {Q_{gain}} = 9555{\text{J}}$ ------- (4)

Step 4: Use the principle of heat lost equals heat gained to find the specific heat of aluminium.
The heat lost by the sphere will be the heat gained by the water.

So we equate the equations (2) and (4) to get, $22 \times {s_a} = 9555$ .
$ \Rightarrow {s_a} = \dfrac{{9555}}{{22}} = 434{\text{Jk}}{{\text{g}}^{ - 1}}^\circ {{\text{C}}^{ - 1}}$

Thus the specific heat of aluminium is ${s_a} = 434{\text{Jk}}{{\text{g}}^{ - 1}}^\circ {{\text{C}}^{ - 1}}$ and the correct option is A.

Note: Here, we assumed that the heat lost to the surroundings of the system is negligible. If this was not the case, then the specific heat of aluminium would be lesser than the obtained value. Also, while substituting values in an equation make sure that the physical quantities are expressed in their S.I. units. If not, then the necessary conversion of units must be done.