Question

A liquid of specific heat $0.3$ at ${90}^o$C is mixed with another liquid of specific heat $0.5$ at ${15}^o$C. If final temperature of the mixture is ${60}^o$C, then find the ratio of masses of the two liquids mixed.

Answer 1

By the Principle of Calorimetry,
Heat lost = Heat gain

$m_1\times 0.3\times (90-60)=m_2\times 0.5\times (60-15)\Rightarrow \frac{m_1}{m_2}=\frac{0.5\times 45}{0.3\times 30}=\frac{5}{2}$