Question

200 g of a solid ball at \[20^\circ C\] is dropped in an equal amount of water at \[80^\circ C\]. The resulting temperature is \[60^\circ \]. This means that specific heat of solid is:
A. One fourth of water
B. One half of water
C. Twice of water
D. Four times of water

Answer 1

Hint: The heat gain by the solid ball is equal to the heat lost by the water. Use the formula for heat lost or gained by the material of mass m and specific heat c to determine the specific heat of the solid ball.

Formula used:
Heat gained or lost by the material of mass m and specific heat c is given as,
\[Q = mc\Delta T\]

Here, \[\Delta T\] is the difference in the temperature.

Complete step by step answer:
We know that the amount of heat gained or lost by the material of mass m and specific heat S is given as,
\[Q = mc\Delta T\]

Here, \[\Delta T\] is the difference in the temperature.

When a solid ball at low temperature is dropped into water at high temperature relative to the ball, the resultant temperature of the system will be greater than the temperature of the solid ball and less than the temperature of water due to loss in heat energy by the water and gain in the heat energy by the solid ball.

According to principle of calorimetry, the heat gained by the solid ball is equal to the heat lost by the water.

Therefore, we can write,
\[{m_b}{s_b}\left( {T - {T_b}} \right) = {m_w}{s_w}\left( {{T_w} - T} \right)\]

Here, \[{m_b}\] is the mass of solid ball, \[{s_b}\] is the specific heat of solid ball, \[{m_w}\] is the mass of water and \[{s_w}\] is the specific heat of water, \[{T_b}\] is the temperature of solid ball, \[{T_w}\] is the initial temperature of water and T is the final temperature of the mixture.

Since the mass of solid ball and water is equal, we can write the above equation as,
\[{s_b}\left( {T - {T_b}} \right) = {s_w}\left( {{T_w} - T} \right)\]
\[ \Rightarrow {s_b} = {s_w}\dfrac{{\left( {{T_w} - T} \right)}}{{\left( {T - {T_b}} \right)}}\]

Substitute \[20^\circ C\] for \[{T_b}\], \[80^\circ C\] for \[{T_w}\] and \[60^\circ C\] for T in the above equation.
 \[{s_b} = {s_w}\dfrac{{\left( {80 - 60} \right)}}{{\left( {60 - 20} \right)}}\]
\[ \Rightarrow {s_b} = \dfrac{1}{2}{s_w}\]

Therefore, the specific heat of a solid ball is one half of water.

So, the correct answer is option (B).

Note:To express the heat lost by the material, take the difference between temperature of material and the final temperature of the mixture and not the difference between final temperature of the mixture and initial temperature of the material. It actually comes from the fact that the sum of heat lost and heat gained is zero.