Chapter 21, Problem 77PQ

(a)

To determine

The work done on the lead block in the process.

Answer to Problem 77PQ

The work done on the lead block in the process is $\underset{\_}{-0.279\text{J}}$.

Explanation of Solution

Given that the mass of the lead block is $10.0\text{kg}$, the temperature of the block is increased from $18.0°\text{C}$ to $55.0°\text{C}$, the pressure remains constant at $1.00\text{atm}$. The coefficient of linear expansion of lead is $28.0\times {10}^{-6}°{\text{C}}^{-1}$.

Write the expression for the work done at constant pressure.

  $W=-P\Delta V$                                                                                                        (I)

Here, $W$ is the work done at constant pressure, $P$ is the pressure, and $\Delta V$ is the change in volume of the block.

Write the expression for the change in the volume of lead block when its temperature changes.

  $\Delta V=\beta V\left(T_f-T_i\right)$                                                                                            (II)

Here, $\beta$ is the coefficient of volume expansion, $V$ is the initial volume of the block, $T_f$ is the final temperature of the block, and $T_i$ is the initial temperature of the block.

Write the expression for the coefficient of volume expansion in terms of the coefficient of linear expansion.

  $\beta =3\alpha$                                                                                                                    (III)

Here, $\alpha$ is the coefficient of linear expansion of the lead block.

Use expression (III) in (II).

  $\Delta V=\left(3\alpha \right)V\left(T_f-T_i\right)$                                                                                       (IV)

Write the expression for the volume of the lead block in terms of its mass and density.

  $V=\frac{m}{\rho }$                                                                                                              (V)

Here, $m$ is the mass of the block, and $\rho$ is the density of the block.

Use expression (V) in (IV).

  $\Delta V=\frac{m}{\rho }\left(3\alpha \right)\left(T_f-T_i\right)$                                                                                 (VI)

Use expression (VI) in (I).

  $W=-P\left[\frac{m}{\rho }\left(3\alpha \right)\left(T_f-T_i\right)\right]$                                                                             (VII)

Conclusion:

Substitute $1.00\text{atm}$ for $P$, $28.0\times {10}^{-6}°{\text{C}}^{-1}$ for $\alpha$, $10.0\text{kg}$ for $m$, $11.3\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $55.0°\text{C}$ for $T_f$, and $18.0°\text{C}$ for $T_i$ in equation (VII) to find $W$.

  $\begin{array}{c}W=-\left(1.00\text{atm}\times \frac{1.013\times {10}^5\text{N}/{\text{m}}^{\text{2}}}{1.00\text{atm}}\right)\left[3\left(28.0\times {10}^{-6}°{\text{C}}^{-1}\right)\left(\frac{10.0\text{kg}}{11.3\times {10}^3\text{kg}/{\text{m}}^{\text{3}}}\right)\left(55.0°\text{C}-18.0°\text{C}\right)\right]\\ =-0.279\text{J}\end{array}$

Therefore, the work done on the lead block when the temperature increases is $\underset{\_}{-0.279\text{J}}$.

(b)

To determine

The amount of energy added to the block by heat in the process.

Answer to Problem 77PQ

The amount of energy added to the block by heat in the process is $\underset{\_}{47.4\text{kJ}}$.

Explanation of Solution

Write the expression for the heat added up into the block while heating.

  $Q=mc\left(T_f-T_i\right)$                                                                                           (VIII)

Here, $Q$ is the energy or heat added to the block, and $c$ is the specific heat capacity of lead.

Conclusion:

Substitute $128\text{J}/\text{kg}\cdot °\text{C}$ for $c$, $10.0\text{kg}$ for $m$, $55.0°\text{C}$ for $T_f$, and $18.0°\text{C}$ for $T_i$ in equation (VIII) to find $Q$.

  $\begin{array}{c}Q=\left(128\text{J}/\text{kg}\cdot °\text{C}\right)\left(10.0\text{kg}\right)\left(55.0°\text{C}-18.0°\text{C}\right)\\ =0.474\times {10}^2\text{J}\\ =0.474\times {10}^5\text{J}\times \frac{1\text{kJ}}{1000\text{J}}\\ =47.4\text{kJ}\end{array}$

Therefore, the amount of energy added to the block by heat is $\underset{\_}{47.4\text{kJ}}$.

(c)

To determine

The change in the internal energy of the block during the process.

Answer to Problem 77PQ

The change in thermal energy of the block during the process is $\underset{\_}{47.4\text{kJ}}$.

Explanation of Solution

The change in thermal energy of the block is equal to the sum of heat added up to the system and the work done.

Write the expression for the change in thermal energy of the block.

  $\Delta E_{\text{th}}=Q+W$                                                                                                (IX)

Here, $\Delta E_{\text{th}}$ is the change in the thermal energy of block.

Conclusion:

Substitute $47.4\text{kJ}$ for $Q$, and $-0.279\text{J}$ for $W$ in equation (IX) to find $\Delta E_{\text{th}}$.

  $\begin{array}{c}\Delta E_{\text{th}}=47.4\text{kJ}-0.279\text{J}\times \frac{1\text{kJ}}{1000\text{J}}\\ =\text{47}\text{.4}\text{kJ}\end{array}$

Therefore, the change in thermal energy of the block during the process is $\underset{\_}{47.4\text{kJ}}$.