Chapter 6, Problem 125QP

Interpretation Introduction

Interpretation:

The specific heat of pipe 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 125QP

Solution:

The specific heat of pipe is found to be $0.450\text{ J/g}°\text{C}$ .

Explanation of Solution

Given Information: Mass of pipe and water is $175\text{ g}$ and $100\text{ g}$ , respectively. The temperatures of pipe and water are $78.24°\text{C}$ and $25.00°\text{C}$ , respectively. The final temperature of the mixture is $33.43°\text{C}$ .

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

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

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

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

Substitute $175\text{ g}$ for $m_{\text{pipe}}$ and $\left(T°\text{C}-78.24°\text{C}\right)$ for $\Delta T_{\text{Cu}}$ in the above equation. Here, $T°\text{C}$ is the temperature of pipe and water at thermal equilibrium.

$q_{\text{pipe}}=150\text{ g}\times C_{\text{pipe}}\times \left(T°\text{C}-78.24°\text{C}\right)$

Substitute $33.43°\text{C}$ for $T°\text{C}$ in the above equation.

$\begin{array}{c}{q}_{\text{Cu}}=175\text{ g}\times C_{\text{pipe}}\times \left(33.43°\text{C}-78.24°\text{C}\right)\\ =175\text{ g}\times C_{\text{pipe}}\times \left(-44.81°\text{C}\right)\\ \text{=}\left(-7841.75\text{ g}°\text{C}\right)C_{\text{pipe}}\end{array}$

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 temperature of the water.

Substitute $\text{100 g}$ for $m_{\text{water}}$ , $\text{4}\text{.184 J/g}°\text{C}$ for $C_{\text{water}}$ , and $\left(T°\text{C}-25.00°\text{C}\right)$ for $\Delta T_{\text{water}}$ in the above equation. Here, $T°\text{C}$ is the temperature of water and pipe at thermal equilibrium.

$q_{\text{water}}=100\text{ g}\times 4.184\text{ J/g}°\text{C}\times \left(T°\text{C}-25.00°\text{C}\right)$

Substitute $33.43°\text{C}$ for $T°\text{C}$ in the above equation.

$\begin{array}{c}{q}_{\text{water}}=100\text{ g}\times 4.184\text{ J/g}°\text{C}\times \left(33.43°\text{C}-25.00°\text{C}\right)\\ =100\text{ g}\times 4.184\text{ J/g}°\text{C}\times 8.43°\text{C}\\ \text{=3527}\text{.12 J}\end{array}$

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

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

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.

$\begin{array}{c}3527.12\text{ J}+\left(-7841.75\text{ g}°\text{C}\right)C_{\text{pipe}}=0\\ \left(-7841.75\text{ g}°\text{C}\right)C_{\text{pipe}}=-3527.12\text{ J}\\ C_{\text{pipe}}=\frac{-3527.12\text{ J}}{-7841.75\text{ g}°\text{C}}\\ C_{\text{pipe}}=0.450\text{ J/g}°\text{C}\end{array}$

Therefore, the specific heat of the pipe is $0.450\text{ J/g}°\text{C}$ and using table 6.2, it could be said that the pipe is made up of chromium $\left(s\right)$ .

Conclusion

The specific heat of the pipe is $0.450\text{ J/g}°\text{C}$ and the material that the pipe is made up of is chromium $\left(s\right)$ .