Chapter 11, Problem 18P

In a showdown on the streets of Laredo, the good guy drops a 5.00-g silver bullet at a temperature of 20.0°C into a 100-cm³ cup of water at 90.0°C. Simultaneously, the bad guy drops a 5.00-g copper bullet at the same initial temperature into an identical cup of water. Which one ends the showdown with the coolest cup of water in the West? Neglect any energy transfer into or away from the container.

To determine

Which one ends up with the coolest cup.

Answer to Problem 18P

Solution: The bad guy’s cup will be coolest among the two cups.

Explanation of Solution

Given info: Mass of silver and copper bullets dropped by good and bad guys respectively is 5.00 g, volume of the cup is $100{\text{cm}}^{\text{3}}$,the temperature of bullets is ${20}^{\text{ο}}\text{C}$ and temperature of water is ${90}^{\text{ο}}\text{C}$.

Write the expression for equilibrium.

$m_s{c}_s\left(T-T_s\right)={\rho }_w{V}_w{c}_w\left(T-T_w\right)$

• $m_s$ is the mass of silver bullet
• $c_s$ is the specific heat of silver
• ${\rho }_w$ is the density of water
• $V_w$ is the volume of water cup
• $c_w$ is the specific heat of water
• T is the equilibrium temperature
• $T_s$ and $T_w$ are the temperatures of silver bullet and water respectively.

Re-arrange the above equation to get T.

$T=\frac{m_s{c}_s{T}_s+{\rho }_w{V}_w{c}_w{T}_w}{m_s{c}_s+{\rho }_w{V}_w{c}_w}$

Substitute $234{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}$ for $c_s$, $4186{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}$ for $c_w$, 5.00 g for $m_s$, $1000{\text{kgm}}^{-3}$ for ${\rho }_w$, $100{\text{cm}}^{\text{3}}$ for $V_w$, ${20}^{\text{ο}}\text{C}$ for $T_s$ and ${90}^{\text{ο}}\text{C}$ for $T_w$ in the above equation.

$\begin{array}{c}T=\frac{\left(5.00\text{g}\right)\left(234{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}\right)\left({20}^{\text{ο}}\text{C}\right)+\left(1000{\text{kgm}}^{-3}\right)\left(100{\text{cm}}^{\text{3}}\right)\left(4186{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}\right)\left({90}^{\text{ο}}\text{C}\right)}{\left(5.00\text{g}\right)\left(234{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}\right)+\left(1000{\text{kgm}}^{-3}\right)\left(100{\text{cm}}^{\text{3}}\right)\left(4186{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}\right)}\\ =\frac{\left(5.00\times {10}^{-3}\text{kg}\right)\left(234{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}\right)\left({20}^{\text{ο}}\text{C}\right)+\left(1000{\text{kgm}}^{-3}\right)\left(100\times {10}^{-6}{\text{m}}^{\text{3}}\right)\left(4186{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}\right)\left({90}^{\text{ο}}\text{C}\right)}{\left(5.00\times {10}^{-3}\text{kg}\right)\left(234{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}\right)+\left(1000{\text{kgm}}^{-3}\right)\left(100\times {10}^{-6}{\text{m}}^{\text{3}}\right)\left(4186{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}\right)}\\ ={89.8}^{\text{ο}}\text{C}\end{array}$

Write the expression for equilibrium.

$m_c{c}_c\left(T-T_c\right)={\rho }_w{V}_w{c}_w\left(T-T_w\right)$

• $m_c$ is the mass of copper bullet
• $c_c$ is the specific heat of copper
• ${\rho }_w$ is the density of water
• $V_w$ is the volume of water cup
• $c_w$ is the specific heat of water
• T is the equilibrium temperature
• $T_c$ and $T_w$ are the temperatures of copper bullet and water respectively.

Re-arrange the above equation to get T.

$T=\frac{m_c{c}_c{T}_c+{\rho }_w{V}_w{c}_w{T}_w}{m_c{c}_c+{\rho }_w{V}_w{c}_w}$

Substitute $387{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}$ for $c_c$, $4186{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}$ for $c_w$, 5.00 g for $m_c$, $1000{\text{kgm}}^{-3}$ for ${\rho }_w$, $100{\text{cm}}^{\text{3}}$ for $V_w$, ${20}^{\text{ο}}\text{C}$ for $T_c$ and ${90}^{\text{ο}}\text{C}$ for $T_w$ in the above equation.

$\begin{array}{c}T=\frac{\left(5.00\text{g}\right)\left(387{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}\right)\left({20}^{\text{ο}}\text{C}\right)+\left(1000{\text{kgm}}^{-3}\right)\left(100{\text{cm}}^{\text{3}}\right)\left(4186{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}\right)\left({90}^{\text{ο}}\text{C}\right)}{\left(5.00\text{g}\right)\left(387{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}\right)+\left(1000{\text{kgm}}^{-3}\right)\left(100{\text{cm}}^{\text{3}}\right)\left(4186{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}\right)}\\ =\frac{\left(5.00\times {10}^{-3}\text{kg}\right)\left(387{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}\right)\left({20}^{\text{ο}}\text{C}\right)+\left(1000{\text{kgm}}^{-3}\right)\left(100\times {10}^{-6}{\text{m}}^{\text{3}}\right)\left(4186{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}\right)\left({90}^{\text{ο}}\text{C}\right)}{\left(5.00\times {10}^{-3}\text{kg}\right)\left(387{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}\right)+\left(1000{\text{kgm}}^{-3}\right)\left(100\times {10}^{-6}{\text{m}}^{\text{3}}\right)\left(4186{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}\right)}\\ ={89.7}^{\text{ο}}\text{C}\end{array}$

Conclusion:

The equilibrium temperature corresponding to the good guy’s cup is slightly greater than that of the bad guy’s cup. Therefore, the bad guy’s cup copper bullet wins as it is the coolest among the two cups