Chapter SH, Problem 5.8APP

To determine

To calculate: Specific heat of the object.

Answer to Problem 5.8APP

Specific heat of the object is $\underset{\_}{-2967.4\text{ J/kg}\cdot °\text{C}}$ .

Explanation of Solution

Given:

Mass of water = $\text{1}\text{.0 kg}$

Specific heat of water = $4184\text{ J/}\left(\text{kg}\cdot °\text{C}\right)$

Mass of object = $4.23\text{ kg}$

Temperature change in water = $3.0°\text{}\text{C}$

Temperature change of the object = $-1.0°\text{}\text{C}$

Formula used:

Change in thermal energy is given by,

  $Q=m\left(T_{\text{f}}-T_{\text{i}}\right)C$

where $m$ is mass, $\left(T_{\text{f}}-T_{\text{i}}\right)$ is temperature change and C is specific heat.

Calculation:

Change in thermal energy of water is calculated as:

  $\begin{array}{l}{Q}_w=m_w{\left(T_{\text{f}}-T_{\text{i}}\right)}_w{C}_w\\ Q_w=\left(\text{1}\text{.0 kg}\right)\left(3.0°\text{C}\right)\left(4184\text{ J/}\left(\text{kg}\cdot °\text{C}\right)\right)\\ Q_w=12552\text{ J}\end{array}$

Change in thermal energy of the object is calculated as:

  $\begin{array}{l}{Q}_o=m_o{\left(T_{\text{f}}-T_{\text{i}}\right)}_o{C}_o\\ Q_o=\left(\text{4}\text{.23 kg}\right)\left(-1.0°\text{C}\right)C_o\\ Q_o=-4.23C_o\text{ kg}\cdot °\text{C}\end{array}$

Since the amount of heat lost is equal to the amount of heat gained,

  $\begin{array}{l}{Q}_w=Q_0\\ 12552\text{ J}=-4.23C_o\text{ kg}\cdot °\text{C}\\ C_o=\frac{12552\text{ J}}{-4.23\text{ kg}\cdot °\text{C}}\\ C_o=-2967.4\text{ J/kg}\cdot °\text{C}\end{array}$

Conclusion:

Specific heat of the object is $\underset{\_}{-2967.4\text{ J/kg}\cdot °\text{C}}$ .