Chapter 15, Problem 73P

(a)

To determine

The entropy change in the universe.

Answer to Problem 73P

The change in entropy of universe is $0.90\text{J/K}$.

Explanation of Solution

The mass of the first block of iron is $0.500\text{kg}$ and it is at temperature $60.0\text{°C}$. The mass of the second block of iron is $0.500\text{kg}$ and it is at temperature $20.0\text{°C}$. The common temperature after the blocks come into contact is $40.0\text{°C}$.

Write the formula for the change in entropy of a system.

$\Delta S=\frac{Q}{T}$

Here, $\Delta S$ is the change in entropy, $Q$ is the heat flow into the system, $T$ is the temperature.

The change in entropy of the universe is the sum of change of entropy of the two blocks.

$\Delta S_U=\Delta S_1+\Delta S_2$

Here, $\Delta S_U$ is the change in entropy of the universe, $\Delta S_1$ is the change in entropy of the first block, $\Delta S_2$ is the change in entropy of the second bag.

Re-write the above equation by substituting the individual expressions.

$\Delta S_U=\frac{Q_1}{T_1}+\frac{Q_2}{T_2}$

Here, $Q_1$ is the heat flow into the first block, $Q_2$ is the heat flow into the second block, $T_1$ is the average temperature of the first block, $T_2$ is the average temperature of the second block.

Re-write the above equation by substituting the individual expressions.

$\Delta S_U=\frac{m_{1}c\Delta T_1}{T_1}+\frac{m_{2}c\Delta T_2}{T_2}$

Here, $m_1$ is the mass of the first block, $m_2$ is the mass of the second block, $\Delta T_1$ is the change in temperature of the first block, $\Delta T_2$ is the change in temperature of the second block, $c$ is the specific heat capacity of the iron.

Conclusion:

Substitute $0.500\text{kg}$ for $m_1$, $0.500\text{kg}$ for $m_2$, $0.44\times {10}^3\text{J/kg}\cdot \text{K}$ for $c$, $40.0\text{°C-}20.0\text{°C}$ for $\Delta T_1$, $40.0\text{°C-}60.0\text{°C}$ for $\Delta T_2$, $\left(40.0\text{°C+}20.0\text{°C}\right)/2$ for $T_1$, $\left(40.0\text{°C+}60.0\text{°C}\right)/2$ for $T_2$

$\begin{array}{c}\Delta S_U=\left(\left(0.500\text{kg}\right)\left(0.44\times {10}^3\text{J/kg}\cdot \text{K}\right)\right)\left[\frac{\left(40.0\text{°C-}20.0\text{°C}\right)}{\left(40.0\text{°C+}20.0\text{°C}\right)/2}+\frac{\left(40.0\text{°C-}60.0\text{°C}\right)}{\left(40.0\text{°C+}60.0\text{°C}\right)/2}\right]\\ =\left(\left(0.500\text{kg}\right)\left(0.44\times {10}^3\text{J/kg}\cdot \text{K}\right)\right)\left[\frac{\left(40.0\text{°C-}20.0\text{°C}\right)}{30.0+273.15\text{K}}+\frac{\left(40.0\text{°C-}60.0\text{°C}\right)}{50.0+273.15\text{K}}\right]\\ =0.90\text{J/K}\end{array}$

The change in entropy of universe is $0.90\text{J/K}$.

(b)

To determine

The entropy change in the universe.

Answer to Problem 73P

The change in entropy of universe is $-2.7\text{J/K}$.

Explanation of Solution

The mass of the first block of iron is $0.500\text{kg}$ and it is at temperature $60.0\text{°C}$. The mass of the second block of iron is $0.500\text{kg}$ and it is at temperature $20.0\text{°C}$. After the two blocks come into contact, the temperature of the first block increases to $80.0\text{°C}$ and the temperature of the second block decreases to $0.0\text{°C}$.

Write the formula for the change in entropy of a system.

$\Delta S=\frac{Q}{T}$

Here, $\Delta S$ is the change in entropy, $Q$ is the heat flow into the system, $T$ is the temperature.

The change in entropy of the universe is the sum of change of entropy of the two blocks.

$\Delta S_U=\Delta S_1+\Delta S_2$

Here, $\Delta S_U$ is the change in entropy of the universe, $\Delta S_1$ is the change in entropy of the first block, $\Delta S_2$ is the change in entropy of the second bag.

Re-write the above equation by substituting the individual expressions.

$\Delta S_U=\frac{Q_1}{T_1}+\frac{Q_2}{T_2}$

Here, $Q_1$ is the heat flow into the first block, $Q_2$ is the heat flow into the second block, $T_1$ is the average temperature of the first block, $T_2$ is the average temperature of the second block.

Re-write the above equation by substituting the individual expressions.

$\Delta S_U=\frac{m_{1}c\Delta T_1}{T_1}+\frac{m_{2}c\Delta T_2}{T_2}$

Here, $m_1$ is the mass of the first block, $m_2$ is the mass of the second block, $\Delta T_1$ is the change in temperature of the first block, $\Delta T_2$ is the change in temperature of the second block, $c$ is the specific heat capacity of the iron.

Conclusion:

Substitute $0.500\text{kg}$ for $m_1$, $0.500\text{kg}$ for $m_2$, $0.44\times {10}^3\text{J/kg}\cdot \text{K}$ for $c$, $0.0\text{°C-}20.0\text{°C}$ for $\Delta T_1$, $80.0\text{°C-}60.0\text{°C}$ for $\Delta T_2$, $\left(0.0\text{°C+}20.0\text{°C}\right)/2$ for $T_1$, $\left(80.0\text{°C+}60.0\text{°C}\right)/2$ for $T_2$

$\begin{array}{c}\Delta S_U=\left(\left(0.500\text{kg}\right)\left(0.44\times {10}^3\text{J/kg}\cdot \text{K}\right)\right)\left[\frac{\left(0.0\text{°C-}20.0\text{°C}\right)}{\left(0.0\text{°C+}20.0\text{°C}\right)/2}+\frac{\left(80.0\text{°C-}60.0\text{°C}\right)}{\left(80.0\text{°C+}60.0\text{°C}\right)/2}\right]\\ =\left(\left(0.500\text{kg}\right)\left(0.44\times {10}^3\text{J/kg}\cdot \text{K}\right)\right)\left[\frac{\left(\text{-}20.0\text{°C}\right)}{10.0+273.15\text{K}}+\frac{\left(20.0\text{°C}\right)}{70.0+273.15\text{K}}\right]\\ =-2.7\text{J/K}\end{array}$

The change in entropy of universe is $-2.7\text{J/K}$.