Question

A liquid of mass m and specific heat S is at a temperature \[2t\]. If another liquid of equal mass and thermal capacity \[1.5\;S\], at a temperature of \[\dfrac{t}{3}\] is added to it, the resultant temperature will be:
A. \[\dfrac{4}{3}t\\ \]
B. \[t \\ \]
C. \[\dfrac{t}{2}\\ \]
D. \[\dfrac{2}{3}t\]

Answer 1

Hint:
The above problem can be resolved using the concepts and fundamentals of the thermal equilibrium. In the thermal equilibrium of liquid, the heat released by any liquid is exactly equal to the amount of heat received by any other liquid. Therefore, the basic formula for the heat lost and the heat gain is applied, and upon comparing both the equations, the desired value of net temperature is obtained.

Complete step by step solution
Given:
The mass of one liquid is, \[m\]
The specific heat of the first liquid is, \[S\].
The temperature of the first liquid is, \[{t_1} = 2t\].
The thermal capacity of another liquid is, \[{S_1} = 1.5\;{\rm{S}}\].
The temperature of the second liquid is, \[{t_2} = \dfrac{t}{3}\].
As the heat loss by one liquid is equal to the heat received by another liquid.
Then the heat loss by one liquid is,
\[{q_1} = {m_1}S\left( {2t - T} \right).............................................\left( 1 \right)\]
Here, T is the resultant temperature.

And the heat received by another liquid is,
\[{q_2} = {m_2}\left( {1.5{\rm{S}}} \right)\left( {T - \dfrac{t}{3}} \right)\]……………………………………. (2)
As the masses are the same, then \[{m_1} = {m_2} = m\].
Solve by comparing the equation 1 and 2 as,
\[\begin{array}{l}
{q_1} = {q_2}\\
{m_1}S\left( {2t - T} \right) = {m_2}\left( {1.5{\rm{S}}} \right)\left( {T - \dfrac{t}{3}} \right)\\
mS\left( {2t - T} \right) = m\left( {1.5{\rm{S}}} \right)\left( {T - \dfrac{t}{3}} \right)\\
2t - T = 3\left( {\dfrac{T}{2}} \right) - \dfrac{t}{2}
\end{array}\]
Further solve the above equation as,
\[\begin{array}{l}
2t - T = 3\left( {\dfrac{T}{2}} \right) - \dfrac{t}{2}\\
5\left( {\dfrac{T}{2}} \right) = 5\left( {\dfrac{t}{2}} \right)\\
T = t
\end{array}\]
Therefore, the resultant temperature will be t and option (B) is correct.

Note:
To solve the given problem, one must go through the concepts and practical application of the heat transfer. Generally, it is observed that heat is being transferred from the higher temperature region to the lower temperature region. Moreover, when the two liquids maintain thermal equilibrium in the context of one another, then the thermal energy interaction restricts within them.