Question

A liquid of mass m and specific heat S is heated to a temperature T. Another liquid of mass $\dfrac{m}{2}$ and specific heat 2S is heated to a temperature 2T. If these two liquids are mixed, the resultant temperature of the mixture:
A. $\dfrac{2}{3}T$
B. $\dfrac{8}{5}T$
C. $\dfrac{3}{3}T$
D. $\dfrac{3}{2}T$

Answer 1

Recall the expression for heat transferred. Since, in the given question we are discussing a condition in which the heat is conserved, the heat lost by one liquid should be the heat gained by the other. Substituting all the given values for mass, specific heat and temperature in the expression for heat transferred and equating the heat transferred for both liquids we get the resultant temperature in terms of T.

Formula used: Expression for heat transferred,
$Q=mc\Delta T$

Complete step by step answer:
We know that heat capacity (C) of a substance is the heat required to raise the temperature of a substance by 1°. So, if we represent Q as the amount of heat required to raise the temperature by ∆T, then,
$Q=C\Delta T$
In the case of specific heat capacity (c) we specify the mass of the substance, that is, it is the heat required to raise the temperature of a unit gram of a substance by unit degree. Therefore,
$Q=mc\Delta T$………………. (1)
In the given case, since the heat is conserved, heat gained by one liquid should be equal to the heat lost by the other when they form the mixture. That is, ∆T should be the same for both liquids.
We are given the following details in the question,
Mass of first liquid=m
Specific heat of first liquid=S
Temperature of first liquid before mixing=T
Mass of second liquid=$\dfrac{m}{2}$
Specific heat of second liquid=2S
Temperature of second liquid before mixing=2T
Let the temperature of both the liquids after mixing be T’, then, we know that the heat lost by liquid at temperature 2T is equal to that gained by liquid with temperature T. That is,
From (1) Heat lost by liquid with temperature 2T is given by,
$Q=m\times S\times (2T-T')$ ………….. (2)
Heat gained by liquid at temperature T is given by,
$-Q=\dfrac{m}{2}\times 2S\times (T-T')$ ………………….. (3)
Adding (2) and (3) we get,
$(m\times S\times (2T-T'))+\left( \dfrac{m}{2}\times 2S\times \left( T-T' \right) \right)=0$
$\left( m\times S\times T' \right)+\left( \dfrac{m}{2}\times 2S\times T' \right)=(m\times S\times 2T)+\left( \dfrac{m}{2}\times 2S\times T \right)$
$T'=\dfrac{(m\times S\times 2T)+\left( \dfrac{m}{2}\times 2S\times T \right)}{\left( m\times S \right)+\left( \dfrac{m}{2}\times 2S \right)}$
$T'=\dfrac{3mST}{2mS}=\dfrac{3}{2}T$
So, we get the resultant temperature of the mixture as$\dfrac{3}{2}T$ .

So, the correct answer is “Option D”.

Note: Though this lengthy explanation gives you conceptual clarity, there is an alternative method to solve this problem by simple substitution. If you know that the resultant temperature after mixing the two liquids is their average temperature, that is,
$T=\dfrac{{{m}_{1}}{{S}_{1}}{{T}_{1}}+{{m}_{2}}{{S}_{2}}{{T}_{2}}}{{{m}_{1}}{{S}_{1}}+{{m}_{2}}{{S}_{2}}}$
Then, you could simply substitute the given values and then get the answer.