Chapter 14, Problem 97P

(a)

To determine

The increasing temperature of the bullet.

Answer to Problem 97P

The increasing temperature of the bullet is $180°\text{C}$.

Explanation of Solution

Write the equation for heat energy required to produce temperature change in a system.

$Q=mc\Delta T$ (I)

Here, $Q$ is the heat energy, $m$ is the mass of the iron bullet, $c$ is the specific heat capacity of the iron material, and $\Delta T$ is the increasing temperature of the bullet.

Write the equation for kinetic energy of system.

$E=\frac{1}{2}m{v}^2$ (II)

Here, $E$ is the kinetic energy of the iron bullet and $v$ is the speed of the bullet.

Conclusion:

Compare the equation (I) and (II) and solve for $\Delta T$.

$\begin{array}{c}mc\Delta T=\frac{1}{2}m{v}^2\\ c\Delta T=\frac{1}{2}{v}^2\\ \Delta T=\frac{v^2}{2c}\end{array}$

Substitute $4.00\times {10}^2\text{m/s}$ for $v$ and $0.44\text{kJ/kg}\cdot \text{K}$ for $c$ in above equation to find $\Delta T$.

$\begin{array}{c}\Delta T=\frac{{\left(4.00\times {10}^2\text{m/s}\right)}^2}{2\left(0.44\text{kJ/kg}\cdot \text{K}\right)\left(\frac{{10}^3\text{J}}{1\text{kJ}}\right)}\\ =\frac{{\left(4.00\times {10}^2\text{m/s}\right)}^2}{2\left(0.44\times {10}^3\text{J/kg}\cdot \text{K}\right)}\\ =182°\text{C}\\ \simeq \text{180}°\text{C}\end{array}$

Therefore, the increasing temperature of the bullet is $180°\text{C}$.

(b)

To determine

Find the equilibrium temperature.

Answer to Problem 97P

The equilibrium temperature is $20.9°\text{C}$.

Explanation of Solution

Write the equation for heat energy required to produce temperature change in iron bullet.

$Q_{\text{Fe}}=m_{\text{Fe}}{c}_{\text{Fe}}\Delta T$ (III)

Here, $Q_{\text{Fe}}$ is the heat energy acquired by the bullet, $m_{\text{Fe}}$ is the mass of the iron bullet, $c$ is the specific heat capacity of the iron material, and $\Delta T$ is the increasing temperature of the bullet.

Write the equation for heat energy required to produce temperature change in wooden block.

$Q_{\text{W}}=m_{\text{W}}{c}_{\text{W}}\Delta T$ (IV)

Here, $Q_{\text{W}}$ is the heat energy acquired by the wooden block, $m_{\text{W}}$ is the mass of the wooden block, $c_{\text{W}}$ is the specific heat capacity of the wooden block, and $\Delta T$ is the increasing temperature bullet and block system.

Write the equation for kinetic energy of system.

$E_{\text{Fe}}=\frac{1}{2}{m}_{\text{Fe}}{v}^2$ (V)

Here, $E_{\text{Fe}}$ is the kinetic energy of the iron bullet and $v$ is the speed of the bullet.

The change in internal energy of the bullet and block system is equal to the initial kinetic energy of the bullet so that $\Delta U=E$.

$\begin{array}{c}\Delta U=Q_{\text{Fe}}+Q_{\text{W}}\\ \frac{1}{2}{m}_{\text{Fe}}{v}^2=m_{\text{Fe}}{c}_{\text{Fe}}\Delta T+m_{\text{W}}{c}_{\text{W}}\Delta T\\ \frac{1}{2}{m}_{\text{Fe}}{v}^2=\left(m_{\text{Fe}}{c}_{\text{Fe}}+m_{\text{W}}{c}_{\text{W}}\right)\Delta T\end{array}$

Replace $T_{\text{f}}-T_{\text{i}}$ for $\Delta T$ and solve the relation for $T_{\text{f}}$.

$\begin{array}{c}\frac{1}{2}{m}_{\text{Fe}}{v}^2=\left(m_{\text{Fe}}{c}_{\text{Fe}}+m_{\text{W}}{c}_{\text{W}}\right)\left(T_{\text{f}}-T_{\text{i}}\right)\\ \frac{1}{2}{m}_{\text{Fe}}{v}^2+T_{\text{i}}=\left(m_{\text{Fe}}{c}_{\text{Fe}}+m_{\text{W}}{c}_{\text{W}}\right)T_{\text{f}}\\ T_{\text{f}}=\frac{m_{\text{Fe}}{v}^2}{2\left(m_{\text{Fe}}{c}_{\text{Fe}}+m_{\text{W}}{c}_{\text{W}}\right)}+T_{\text{i}}\end{array}$ (VI)

Conclusion:

Substitute $4.00\times {10}^2\text{m/s}$ for $v$, $10.0\text{g}$ for $m_{\text{Fe}}$, $0.44\text{kJ/kg}\cdot \text{K}$ for $c_{\text{Fe}}$, $20.0°\text{C}$ for $T_{\text{i}}$, $0.500\text{kg}$ for $m_{\text{W}}$, and $\text{1680}\text{J/kg}\cdot \text{K}$ for $c_{\text{W}}$ in equation (VI) to find $T_{\text{f}}$.

$\begin{array}{c}{T}_{\text{f}}=\frac{\left(10.0\text{g}\right)\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right){\left(4.00\times {10}^2\text{m/s}\right)}^2}{2\left[\left(10.0\text{g}\right)\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\left(0.44\text{kJ/kg}\cdot \text{K}\right)\left(\frac{{10}^3\text{J}}{1\text{kJ}}\right)+\left(0.500\text{kg}\right)\left(\text{1680}\text{J/kg}\cdot \text{K}\right)\right]}+20.0°\text{C}\\ =\frac{\left(10.0\times {10}^{-3}\text{kg}\right){\left(4.00\times {10}^2\text{m/s}\right)}^2}{2\left[\left(10.0\times {10}^{-3}\text{kg}\right)\left(0.44\times {10}^3\text{J/kg}\cdot \text{K}\right)+\left(0.500\text{kg}\right)\left(\text{1680}\text{J/kg}\cdot \text{K}\right)\right]}+20.0°\text{C}\\ =\frac{1600\text{kg}\cdot {\text{m}}^2/{\text{s}}^2}{2\left[\left(4.4\text{J/K}\right)+840\text{J/K}\right]}+20.0°\text{C}\\ =20.9°\text{C}\end{array}$

Therefore, the equilibrium temperature is $20.9°\text{C}$.