Chapter 12.1, Problem 5PP

To determine

To Calculate: The specific heat of the metal in the weights.

Answer to Problem 5PP

  $0.253\text{J/g}\cdot °\text{C}$

Explanation of Solution

Given:

Mass of metal, $m_1=1.00\times {10}^2\text{g}$

Initial temperature metal, $T_1=100.0°\text{C}$

Mass of water, $m_2=1.00\times {10}^2\text{g}$

Initial temperature of water, $T_2=35.0°\text{C}$

Final temperature of the mixture, $T=45.0°\text{C}$

Formula Used:

  $Q=mc\Delta T$

Where, m is the mass, c is the specific heat capacity and $\Delta T$ is the difference in temperature.

Calculations:

Specific heat capacity of water is $c=4.18\text{J/g}\cdot \text{°C}$

Let the specific heat of the metal be $c_m$ .

Heat lost by metal = Heat gained by water

  $\begin{array}{c}-Q_1=Q_2\\ 3m_1{c}_m\left(T_1-T\right)=m_{2}c\left(T-T_2\right)\\ c_m=\frac{m_{2}c\left(T-T_2\right)}{3m_1\left(T_1-T\right)}\\ =\frac{1.00\times {10}^2\text{kg}\times 4.18\text{J/g}\cdot °\text{C}\times \left(45°\text{C}-35°\text{C}\right)}{3\times 1.00\times {10}^2\text{kg}\left(100°\text{C}-45°\text{C}\right)}\\ =0.253\text{J/g}\cdot °\text{C}\end{array}$

Conclusion:

Thus, the specific heat of the metal in the weights $0.253\text{J/g}\cdot °\text{C}$ .