Chapter 12, Problem 52A

To determine

Final temperature of water and zinc.

Answer to Problem 52A

  $T_f=12.7°\text{C}$

Explanation of Solution

Given:

The given figure is shown below.

  

Mass of zinc, $m_1=10\text{kg}$

Initial temperature of zinc, $T_{1i}=71°\text{C}$

Mass of water, $m_2=20\text{kg}$

Initial temperature of water, $T_{2i}=10°\text{C}$

Final temperature of zinc and water is same, $T_{1f}=T_{2f}$

Formula Used:

Specific heat formula,

  $Q=mC\Delta T$

Here, $Q$ is heat energy, $m$ is mass, $C$ is specific heat, and $\Delta T$ is change in temperature.

From the law of thermodynamics,

  $Q_1+Q_2=0$

Here $Q_1$ is heat energy of water and $Q_2$ is heat energy of zinc.

Calculation:

As specific heat of water is $C_2=4180\text{J/kg}\cdot \text{K}$ and specific heat of zinc is $C_1=388\text{J/kg}\cdot \text{K}$

And $T_{1f}=T_{2f}=T_f$

From the above equation,

  $\begin{array}{c}{m}_1{C}_1\left(T_{1f}-T_{1i}\right)+m_2{C}_2\left(T_{2f}-T_{2i}\right)=0\\ T_f=\frac{m_1{C}_1{T}_{1i}+m_2{C}_2{T}_{2i}}{m_1{C}_1+m_2{C}_2}\\ T_f=\frac{\left(10\times 388\times 71\right)+\left(20\times 4180\times 10\right)}{\left(10\times 388\right)+\left(20\times 4180\right)}\\ =12.7°\text{C}\end{array}$

Conclusion:

Therefore, final temperature of water and zinc is $12.7°\text{C}$ .