Chapter 14, Problem 118P

(a)

To determine

Whether the blocks are necessarily in physical contact.

Answer to Problem 118P

No, the blocks need not necessarily be in physical contact with each other.

Explanation of Solution

For the two blocks to be in thermal contact, the blocks need not touch each other. They can transfer heat by convection and radiation. Other method to be in thermal equilibrium is to connect a bridge of high conducting material between them. Therefore, the blocks need not necessarily be in physical contact with each other.

(b)

To determine

Whether the blocks will have the same internal energy if the temperatures of the blocks are same.

Answer to Problem 118P

The blocks should have the same mass to have the same internal energy.

Explanation of Solution

The blocks are made up of the same material, aluminum. They have the same internal energy and the material of both blocks are the same. But for the internal energy of the blocks to become same, the blocks should have the same mass. This is the necessary condition for the internal energy of the blocks to be same. Therefore, the blocks should have the same mass to have the same internal energy.

(c)

To determine

Whether there is a net energy transfer between the blocks if their internal energies are not equal.

Answer to Problem 118P

No, there is not net transfer of energy even though their internal energies are not equal.

Explanation of Solution

Energy transfer is associated with difference in temperature. The blocks are kept at the same temperatures. There is no difference in the temperature which does not owe to the transfer of heat energy. Therefore, there is not net transfer of energy even though their internal energies are not equal.

(d)

To determine

The final equilibrium temperature and the change in internal energy of cold and hot block.

Answer to Problem 118P

The final equilibrium temperature and the change in internal energy of cold and hot block are $\underset{\_}{25.0°\text{C}}$, $\underset{\_}{13.5\text{kJ}}$ and $\underset{\_}{-13.5\text{kJ}}$ respectively.

Explanation of Solution

Write the expression of the total transfer associated with the blocks.

$Q_{\text{hot block cooling}}=Q_{\text{cold block warming}}$                                                                                       (I)

Write the expression for the heat associated with the cold block.

$Q_{\text{cold block warming}}=m_2{c}_{Al}\left(T-T_{\text{cool}}\right)$ (II)

Here, $m_2$ is the mass of the cold block, $c_{Al}$ is the specific heat of Aluminum, $T$ is the equilibrium temperature and $T_{\text{cool}}$ is the temperature of the cold block.

Write the expression for the heat associated with the hot block.

$Q_{\text{hot block cooling}}=m_1{c}_{Al}\left(T-T_{\text{hot}}\right)$ (III)

Here, $m_1$ is the mass of the hot block, $T$ is the equilibrium temperature and $T_{\text{hot}}$ is the temperature of the hot block.

Conclusion:

Substitute $m_1{c}_{Al}\left(T-T_{\text{hot}}\right)$ for $Q_{\text{hot block cooling}}$ and $m_2{c}_{Al}\left(T-T_{\text{cool}}\right)$ for $Q_{\text{cold block warming}}$ in the equation (I).

$m_1{c}_{Al}\left(T-T_{\text{hot}}\right)=m_2{c}_{Al}\left(T-T_{\text{cool}}\right)$

Substitute $1.00\text{kg}$ for $m_1$, $900\text{J}/\text{kg}\cdot \text{K}$ for $c_{Al}$, $40.0°\text{C}$ for $T_{\text{hot}}$, $3.00\text{kg}$ for $m_2$ and $20.0°\text{C}$ for $T_{\text{cool}}$ in the above equation and rearrange it to find the equilibrium temperature.

$\begin{array}{c}\left(1.00\text{kg}\right)\left(900\text{J}/\text{kg}\cdot \text{K}\right)\left(T-40.0°\text{C}\right)=\left(3.00\text{kg}\right)\left(900\text{J}/\text{kg}\cdot \text{K}\right)\left(T-20.0°\text{C}\right)\\ 40.0°\text{C}+60.0°\text{C}=T+3.00T\\ T=25.0°\text{C}\end{array}$

Substitute $1.00\text{kg}$ for $m_1$, $900\text{J}/\text{kg}\cdot \text{K}$ for $c_{Al}$, $40.0°\text{C}$ for $T_{\text{hot}}$ and $25.0°\text{C}$ for $T$ in the equation (III).

$\begin{array}{c}{Q}_{\text{hot block cooling}}=\left(1.00\text{kg}\right)\left(900\text{J}/\text{kg}\cdot \text{K}\right)\left(25.0°\text{C}-40.0°\text{C}\right)\\ =-13.5\text{kJ}\end{array}$

Substitute $3.00\text{kg}$ for $m_2$, $900\text{J}/\text{kg}\cdot \text{K}$ for $c_{Al}$, $20.0°\text{C}$ for $T_{\text{cool}}$ and $25.0°\text{C}$ for $T$ in the equation (II).

$\begin{array}{c}{Q}_{\text{cold block warming}}=\left(3.00\text{kg}\right)\left(900\text{J}/\text{kg}\cdot \text{K}\right)\left(25.0°\text{C}-20.0°\text{C}\right)\\ =13.5\text{kJ}\end{array}$

Therefore, the final equilibrium temperature and the change in internal energy of cold and hot block are $\underset{\_}{25.0°\text{C}}$, $\underset{\_}{13.5\text{kJ}}$ and $\underset{\_}{-13.5\text{kJ}}$ respectively.