Chapter 12, Problem 45A

To determine

Specific heat of tungsten.

Answer to Problem 45A

  $C_2=170.6\text{J/kg}\cdot \text{K}$

Explanation of Solution

Given:

Mass of water, $m_1=2\times {10}^2\text{g}=0.2\text{kg}$

Initial temperature of water, $T_{1i}=20°\text{C}$

Mass of tungsten, $m_2=1\times {10}^2\text{g}=0.1\text{kg}$

Initial temperature of tungsten, $T_{2i}=100°\text{C}$

Final temperature of tungsten and water, $T_{1f}=T_{2f}=21.6°\text{C}$

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,

  $Q_1+Q_2=0$

Where $Q_1$ and $Q_2$ is heat energy to different systems.

Calculation:

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

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

From the above equation,

  $\begin{array}{c}{m}_1{C}_1\left(T_{1f}-T_{1i}\right)+m_2{C}_2\left(T_{2f}-T_{2i}\right)=0\\ C_2=-\frac{m_1{C}_1\left(T_{1i}-T_{2i}\right)}{m_2\left(T_{2i}-T_{1i}\right)}\\ C_2=-\frac{0.2\times 4180\times \left(21.6°-20°\right)}{0.1\times \left(21.6°-100°\right)}\\ =170.6\text{J/kg}\cdot \text{K}\end{array}$

Conclusion:

Therefore, specific heat of tungsten is $C_2=170.6\text{J/kg}\cdot \text{K}$