Question

The temperature of equal masses of three different liquids $A$, $B$ and $C$ are ${12}^{0}C$, ${19}^{0}C$ and ${28}^{0}C$ respectively. When $A$ and $B$ are mixed the temperature is ${16}^{0}C$ and when $B$ and $C$ are mixed it is ${23}^{0}C$. The temperature when $A$ and $C$ are mixed is

• A. ${10.1}^{0}C$
• B. ${20.2}^{0}C$
• C. ${30.3}^{0}C$
• D. ${40.4}^{0}C$

Answer 1

Answer: (B)

${20.2}^{0}C$

When A and B are mixed, the common temp becomes ${16}^{0}C$.
Heat lost by B is equal to the heat gained by A.
Given:
$m_A=m_B=m_C=m$
Heat lost by B: $Q_B=m_B{s}_B(19-16)=m{s}_B(19-16)$
Heat gained by A: $Q_A=m_A{s}_A(16-12)=m{s}_A(16-12)$
$\therefore m{s}_B(19-16)=m{s}_A(16-12)$
$\Rightarrow 3s_B=4s_A$
$\therefore 12s_B=16s_A$ .....(1)
When B and C are mixed, the common temp becomes ${23}^{0}C$.
Heat lost by C is equal to the heat gained by B.
Heat lost by C: $Q_C=m_C{s}_C(28-23)=5m{s}_C$
Heat gained by B: $Q_B=m_B{s}_B(23-19)=4m{s}_B$
$\Rightarrow 4s_B=5s_C$
$\therefore 12s_B=15s_C$ .....(2)
From (1) and (2)
$16s_A=15s_C$
$\therefore s_A=\frac{15}{16}{s}_C$
When A and C are mixed, let T be the common temp.
Heat lost by C: $Q_C=m_C{s}_C(28-T)=m{s}_C(28-T)$ .....(3)
Heat gained by A: $Q_B=m_A{s}_A(T-12)=m{s}_A(T-12)=m\frac{15}{16}{s}_C(T-12)$ ......(4)
Equating (3) and (4)
$m{s}_C(28-T)=m\frac{15}{16}{s}_C(T-12)$
$\Rightarrow 16(28-T)=15(T-12)$
$\Rightarrow 31T=16\ast 28+15\ast 12=448+180=628$
$\Rightarrow T={20.25}^{0}C$