Question

Three different substances have their specific heats in the ratio $2:3:1$ and their temperatures are increased in the ratio $3:1:2$ respectively. If same heat is supplied to these substances, find the ratio of their masses.

Answer 1

Let the three substances be A, B, C.

We know,

$Q=mC\Delta T⟹m=\frac{Q}{C\Delta T}$

The ratio becomes,

$m_A:m_B:m_C=\frac{Q}{C_A\Delta T_A}:\frac{Q}{C_B\Delta T_B}:\frac{Q}{C_C\Delta T_C}$, since heat is supplied same, Q gets cancelled.

And also its given, $C_A:C_B:C_C=2:3:1$ and $\Delta T_A:\Delta T_B:\Delta T_C=3:1:2$

So the ratio becomes,

$m_A:m_B:m_C=\frac{1}{2\times 3}:\frac{1}{3\times 1}:\frac{1}{1\times 2}=\frac{1}{6}:\frac{1}{3}:\frac{1}{2}$