Chapter 11, Problem 23P

Equal 0.400-kg masses of lead and tin at 60.0°C are placed in 1.00 kg of water at 20.0°C. (a) What is the equilibrium temperature of the system? (b) If an alloy is half lead and half tin by mass, what specific heat would you anticipate for the alloy? (c) How many atoms of tin N_Sn, are in 0.400 kg of tin, and how many atoms of lead N_Pb are in 0.400 kg of lead? (d) Divide the number N_Sn of tin atoms by the number N_Pb of lead atoms and compare this ratio with the specific heat of tin divided by the specific beat of lead. What conclusion can be drawn?

To determine

The equilibrium temperature of the system.

Answer to Problem 23P

Solution: The equilibrium temperature of the system is ${21.3}^{\text{ο}}\text{C}$.

Explanation of Solution

Given info: Mass of lean and tin is 0.400 kg, temperature of lean and tin is ${60.0}^{\text{ο}}\text{C}$, mass of water is 1.00 kg and temperature of water is ${20.0}^{\text{ο}}\text{C}$.

Write the expression for equilibrium.

$m_w{c}_w\left(T-T_w\right)=m{c}_l\left(T_l-T\right)+m{c}_t\left(T_l-T\right)$

• $m$ is the mass of lead and tin
• $c_l$ is the specific heat of lead
• $c_t$ is the specific heat of tin
• $m_w$ is the mass of water
• $c_w$ is the specific heat of water
• T is the equilibrium temperature
• $T_l$ and $T_t$ are the temperatures of lead and tin respectively
• $T_w$ is the temperature of water

Re-arrange the above equation to get T.

$T=\frac{m{c}_l{T}_l+m{c}_t{T}_t+m_w{c}_w{T}_w}{m{c}_c+m{c}_t+m_w{c}_w}$

Substitute $128{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}$ for $c_l$, $218{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}$ for $c_t$, $4186{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}$ for $c_w$, 0.400 kg for m, 1.00 kg for $m_w$, ${20}^{\text{ο}}\text{C}$ for $T_w$, ${60}^{\text{ο}}\text{C}$ for $T_t$ and ${60}^{\text{ο}}\text{C}$ for $T_l$ in the above equation.

$\begin{array}{c}T=\frac{\left(0.400\text{kg}\right)\left(128{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}\right)\left({60}^{\text{ο}}\text{C}\right)+\left(0.400\text{kg}\right)\left(218{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}\right)\left({60}^{\text{ο}}\text{C}\right)+\left(1.00\text{kg}\right)\left(4186{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}\right)\left({20}^{\text{ο}}\text{C}\right)}{\left(0.400\text{kg}\right)\left(128{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}\right)+\left(0.400\text{kg}\right)\left(218{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}\right)+\left(1.00\text{kg}\right)\left(4186{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}\right)}\\ ={21.28}^{\text{ο}}\text{C}\\ \approx {21.3}^{\text{ο}}\text{C}\end{array}$

Conclusion:

The equilibrium temperature of the system is ${21.3}^{\text{ο}}\text{C}$.

To determine

The specific heat of the alloy.

Answer to Problem 23P

Solution: The specific heat of the alloy is $173{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}$.

Explanation of Solution

Given info: Mass of lean and tin is 0.400 kg, temperature of lean and tin is ${60.0}^{\text{ο}}\text{C}$, mass of water is 1.00 kg and temperature of water is ${20.0}^{\text{ο}}\text{C}$.

The alloy is half lead and half tin by mass. Moreover, the change in temperature of the alloy is same as the change in temperature of lead and tin. Therefore, the specific heat of the alloy is,

$c_a=\frac{c_l+c_t}{2}$

• $c_a$ is the specific heat of the alloy
• $c_t$ is the specific heat of tin
• $c_l$ is the specific heat of lead

Substitute $128{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}$ for $c_l$ and $218{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}$ for $c_t$ in the above equation.

$\begin{array}{c}T=\frac{\left(128{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}\right)+\left(218{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}\right)}{2}\\ =173{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}\end{array}$

Conclusion:

The specific heat of the alloy is $173{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}$.

To determine

The number of atoms of tin and lead.

Answer to Problem 23P

Solution: The number of atoms of tin and lead is $2.03\times {10}^{24}$ and $1.16\times {10}^{24}$.

Explanation of Solution

Given info: Mass of lean and tin is 0.400 kg, temperature of lean and tin is ${60.0}^{\text{ο}}\text{C}$, mass of water is 1.00 kg and temperature of water is ${20.0}^{\text{ο}}\text{C}$.

Write the expression to calculate the number of atoms of tin.

$N_t=N_A\left(\frac{m}{M_t}\right)$

• $N_t$ is the number of atoms of tin
• $N_A$ is the Avogadro number
• $M_t$ is molar mass of tin

Substitute $6.023\times {10}^{23}{\text{mol}}^{-1}$ for $N_A$, 0.400 kg for m,and $118.71\text{g/mol}$ for $M_t$ in the above equation to find $N_t$

$\begin{array}{c}{N}_t=\left(6.023\times {10}^{23}{\text{mol}}^{-1}\right)\left(\frac{0.400\text{kg}}{118.71\text{g/mol}}\right)\\ =\left(6.023\times {10}^{23}{\text{mol}}^{-1}\right)\left(\frac{0.400\times {10}^3\text{g}}{118.71\text{g/mol}}\right)\\ =2.03\times {10}^{24}\end{array}$

Write the expression to calculate the number of atoms in lead.

$N_l=N_A\left(\frac{m}{M_l}\right)$

• $N_l$ is the number of atoms of lead
• $N_A$ is the Avogadro number
• $M_l$ is molar mass of lead

Substitute $6.023\times {10}^{23}{\text{mol}}^{-1}$ for $N_A$, 0.400 kg for m and $207.2\text{g/mol}$ for $M_l$ to find $N_l$.

$\begin{array}{c}{N}_l=\left(6.023\times {10}^{23}{\text{mol}}^{-1}\right)\left(\frac{0.400\text{kg}}{207.2\text{g/mol}}\right)\\ =\left(6.023\times {10}^{23}\right)\left(\frac{0.400\times {10}^3\text{g}}{207.2\text{g/mol}}\right)\\ =1.16\times {10}^{24}\end{array}$

Conclusion:

Therefore, the number of atoms of tin and lead is $2.03\times {10}^{24}$ and $1.16\times {10}^{24}$.

To determine

The ratio of number of atoms and specific heat capacities of tin and lead.

Answer to Problem 23P

Solution: The ratio of number of atoms and specific heat capacities of tin and lead is equal.

Explanation of Solution

Given info: Mass of lean and tin is 0.400 kg, temperature of lean and tin is ${60.0}^{\text{ο}}\text{C}$, mass of water is 1.00 kg and temperature of water is ${20.0}^{\text{ο}}\text{C}$.

The ratio of number of atoms of tin and lead is,

$r_1=\frac{N_t}{N_l}$

Substitute $1.93\times {10}^{24}$ for $N_t$ and $1.16\times {10}^{24}$ for $N_l$ to find $r_l$.

$\begin{array}{c}{r}_l=\frac{1.93\times {10}^{24}}{1.16\times {10}^{24}}\\ =1.66\end{array}$ (I)

The ratio of specific heats of tin and lead is,

$r_2=\frac{c_t}{c_l}$

Substitute $128{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}$ for $c_l$ and $218{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}$ for $c_t$ to find $r_2$ .

$\begin{array}{c}{r}_2=\frac{218{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}}{128{\text{Jkg}}^{-1}{\text{/}}^{\text{ο}}\text{C}}\\ =1.70\end{array}$ (II)

From Equations (I) and (II), the ratio of number of atoms and specific heats of tin and lead is nearly the same.

Conclusion:

The ratio of number of atoms and specific heat capacities of tin and lead is equal