Chapter 6, Problem 126QP

A 150.0-g sample of copper is heated to $\text{89}\text{.3°C}\text{.}$ The copper is then placed in 125.0 g of water held in a calorimeter at $\text{22}\text{.5°C}\text{.}$ The final temperature of mixture is $\text{29}\text{.0°C}\text{.}$ Assuming no heat is lost from the water, What is the specific heat of copper?

Interpretation Introduction

Interpretation:

The specific heat of copper is to be determined.

Concept Introduction:

Specific heat of a substance is defined as the amount of heat required to raise the temperature of $1\text{ g}$ of that substance by $1°\text{C}$ .

Answer to Problem 126QP

Solution:

The specific heat of copper is found to be $0.38\text{ J/g}°\text{C}$ .

Explanation of Solution

Given Information: Mass of copper and water is $150\text{ g}$ and $125\text{ g}$ , respectively. The temperatures of copper and water are $89.3°\text{C}$ and $22.5°\text{C}$ , respectively. The final temperature of the mixture is $29.0°\text{C}$ .

Heat transfer takes place from high temperature to low temperature. So, the temperature of copper will decrease and that of water will increase to attain thermal equilibrium. At thermal equilibrium, temperature of both copper and water will be equal.

The heat lost by the copper can be calculated as follows.

$q_{\text{Cu}}=m_{\text{Cu}}\times C_{\text{Cu}}\times \Delta T_{\text{Cu}}$

Here, $q_{\text{Cu}}$ is the heat lost by the copper, $m_{\text{Cu}}$ is the given mass of the copper, $C_{\text{Cu}}$ is the specific heat of the copper, and $\Delta T_{\text{Cu}}$ is the decrease in temperature of the copper.

Substitute $150\text{ g}$ for $m_{\text{Cu}}$ and $\left(T°\text{C}-89.3°\text{C}\right)$ for $\Delta T_{\text{Cu}}$ in the above equation. Here, $T°\text{C}$ is the temperature of copper and water at thermal equilibrium.

$q_{\text{Cu}}=150\text{ g}\times C_{\text{Cu}}\times \left(T°\text{C}-89.3°\text{C}\right)$

Substitute $29.0°\text{C}$ for $T°\text{C}$ in the above equation.

$\begin{array}{c}{q}_{\text{Cu}}=150\text{ g}\times C_{\text{Cu}}\times \left(29.0°\text{C}-89.3°\text{C}\right)\\ =150\text{ g}\times C_{\text{Cu}}\times \left(-60.3°\text{C}\right)\\ \text{=}\left(-9045g.°\text{C}\right)C_{\text{Cu}}\end{array}$

The heat gained by the water can be calculated as follows.

$q_{\text{water}}=m_{\text{water}}\times C_{\text{water}}\times \Delta T_{\text{water}}$

Here, $q_{\text{water}}$ is the heat gained by the water, $m_{\text{water}}$ is the given mass of the water, $C_{\text{water}}$ is the specific heat of the water, and $\Delta T_{\text{water}}$ is the increase in the temperature of water.

Substitute $\text{125 g}$ for $m_{\text{water}}$ , $\text{4}\text{.184 J/g}°\text{C}$ for $C_{\text{water}}$ , and $\left(T°\text{C}-22.5°\text{C}\right)$ for $\Delta T_{\text{water}}$ in the above equation. Here, $T°\text{C}$ is the temperature of water and copper at thermal equilibrium.

$q_{\text{water}}=125\text{ g}\times 4.184\text{ J/g}°\text{C}\times \left(T°\text{C}-22.5°\text{C}\right)$

Substitute $29.0°\text{C}$ for $T°\text{C}$ in the above equation.

$\begin{array}{c}{q}_{\text{water}}=125\text{ g}\times 4.184\text{ J/g}°\text{C}\times \left(29.0°\text{C}-22.5°\text{C}\right)\\ =125\text{ g}\times 4.184\text{ J/g}°\text{C}\times 6.5°\text{C}\\ \text{=3399}\text{.5 J}\end{array}$

As no heat is lost from the water, according to the law of conservation of energy, the total heat of the system remains conserved.

$q_{\text{water}}+q_{\text{Cu}}=0$

The specific heat of the copper can be calculated using the above equation and the calculated values of $q_{\text{water}}$ and $q_{\text{Cu}}$ are as follows.

$\begin{array}{c}3399.5\text{ J}+\left(-9045\text{ g}°\text{C}\right)C_{\text{Cu}}=0\\ \left(-9045\text{ g}°\text{C}\right)C_{\text{Cu}}=-3399.5\text{ J}\\ C_{\text{Cu}}=\frac{-3399.5\text{ J}}{-9045\text{ g}°\text{C}}\\ C_{\text{Cu}}=0.38\text{ J/g}°\text{C}\end{array}$

Therefore, the specific heat of the copper is $0.38\text{ J/g}°\text{C}$ .

Conclusion

The specific heat of the copper is $0.38\text{ J/g}°\text{C}$ .