Question

0.1 m$^{3}$ of water at $80^{\circ}C$ is mixed with 0.3 m$^{3}$ of water at $60^{\circ}C$. The final temperature of the mixture is

• A. $65^{\circ}C$
• B. $70^{\circ}C$
• C. $60^{\circ}C$
• D. $75^{\circ}C$

Hint:

When mixing the same substance, $t_{final} = \frac{m_1t_1 + m_2t_2}{m_1 + m_2}$.

Answer 1

The Correct Option is A

Step 1: Calorimetry Principle
Heat lost by hot water = Heat gained by cold water. Let $t$ be the final temperature.
Step 2: Energy Equation
$m_1 s(80 - t) = m_2 s(t - 60)$. Since density is constant, volume can be used as a proxy for mass: $0.1(80 - t) = 0.3(t - 60)$.
Step 3: Solve for t
$80 - t = 3t - 180 \Rightarrow 4t = 260 \Rightarrow t = 65^{\circ}C$.
Final Answer: (a)