Question

Two liquids A and B are at 32⁰C and 24⁰C. When mixed in equal masses the temperature of the mixture is found to be 28⁰C. Their specific heats are in the ratio of

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

Answer 1

Answer: (C)

1 : 1

Answer 2

Heat lost by A = Heat gained by B

⇒$m_{A} \times c_{A} \times (T_{A} - T) = m_{B} \times c_{B} \times (T - T_{B})$ Since $m_{A} = m_{B}$

and Temperature of the mixture (T) = 28°C

∴ $c_{A} \times (32 - 28) = c_{B} \times (28 - 24)$ ⇒ $\frac{c_{A}}{c_{B}} = 1:1$