Chapter 12, Problem 75A

To determine

Relative masses of blocks.

Answer to Problem 75A

  $m_{Cu}=2.33m_{Al}$

Explanation of Solution

Given:

Initial temperature of copper block, $T_{1i}=100°\text{C}$

Final temperature of copper block, $T_{1f}=60°\text{C}$

Initial temperature of aluminum block, $T_{2i}=20°\text{C}$

Final temperature of aluminum block, $T_{2f}=60°\text{C}$

Formula Used:

From the law of thermodynamics,

  $Q_1+Q_2=0$

Where $Q_1$ and $Q_2$ is heat energy of two different systems.

Calculation:

Here, $Q_1=Q_{Cu}$ and $Q_2=Q_{Al}$ .

As specific heat of aluminum is $C_{Al}=897\text{J/kg}\cdot \text{K}$ and specific heat of copper is $C_{Cu}=385\text{J/kg}\cdot \text{K}$

So,

  $\begin{array}{l}{Q}_{Cu}+Q_{Al}=0\\ m_{Cu}{C}_{Cu}\Delta T_{Cu}+m_{Al}{C}_{Al}\Delta T_{Al}=0\\ m_{Cu}\times 385\times \left(60°-100°\right)+m_{Al}\times 897\times \left(60°-20°\right)=0\\ \Rightarrow \left(m_{Cu}\times 385\right)+\left(m_{Al}\times 897\right)=0\\ \text{So,}\\ m_{Cu}=\frac{897}{385}\times m_{Al}\\ \Rightarrow m_{Cu}=2.33m_{Al}\end{array}$

Conclusion:

Therefore, relative masses of blocks are $m_{Cu}=2.33m_{Al}$ .