A liquid of mass M and specific heat S is at a temperature 2 t. If another liquid of mass M and thermal capacity 1.5 times, at a temperature of t/3 is added to it, the resultant temperature will be:

Let the common temperature of the mixture be T.

For first liquid, mass, M, specific heat of liquid, S and initial temperature of liquid, 2t.

For second liquid, mass, M, specific heat of liquid, 1.5 S and initial temperature of liquid, $\frac{t}{3}$

Heat loss by the hot liquid = Heat consumed by cold liquid

$\mathrm{MS}(2\mathrm{t}-\mathrm{T})=\mathrm{M}(1.5\mathrm{S})\left(\mathrm{T}-\frac{\mathrm{t}}{3}\right)$

$\Rightarrow 2\mathrm{t}-\mathrm{T}=1.5\mathrm{T}-\frac{\mathrm{t}}{2}$

$\Rightarrow \frac{5}{2}\mathrm{T}=\frac{5}{2}\mathrm{t}$

$\Rightarrow T=t$