Question

Two samples, each containing $400g$ of water, are mixed. The temperature of sample A before mixing was ${20}^{\circ }C$ and that of sample B was ${65}^{\circ }C$. What will be the final temperature of the mixture? (Specific heat of water = 4186 J/kg K)

• A. ${22.5}^{\circ }C$
• B. ${42.5}^{\circ }C$
• C. ${32.5}^{\circ }C$
• D. ${52.5}^{\circ }C$

Answer 1

The correct option is B ${42.5}^{\circ }C$

Given:
Mass of sample A and B, $m_A=m_B=400g=0.4kg$
Temperature of sample A, $T_A={20}^{\circ }C$
Temperature of sample B, $T_B={65}^{\circ }C$
Specific heat of water, $S_w=4186J/kgK$
The heat lost by sample A will be equal to heat gained by sample B.
Let the temperature of the mixture be $T_m$.
Heat lost = Heat gained
$\therefore$ $m_A{S}_w(T_B-T_m)=m_B{S}_w(T_m-T_A)$
$\Rightarrow 0.4\times (65-T_m)=0.4\times (T_m-20)$
$\Rightarrow 2T_m=85$
$\Rightarrow T_m={42.5}^{\circ }C$