Question

A liquid of specific heat $0.3cal/g^{o}C$ at ${90}^{o}C$ is mixed with another liquid of specific heat $0.5cal/g^{o}C$ at ${15}^{o}C$. If the resultant temperature of mixture is ${60}^{o}C$ then the ratio of their masses is:

• A. 5 : 2
• B. 2 : 5
• C. 1 : 1
• D. 3 : 4

Answer 1

Answer: (A)

5 : 2

Heat lost by one liquid $=$ Heat gained by another liquid
$m_1.\left(0.3\right)(90-60)=m_2.\left(0.5\right)(60-15)$
$\frac{m_1}{m_2}=5/2$