Chapter 21, Problem 29PQ

(a)

To determine

The amount of heat transferred in order for the whole process to take place.

Answer to Problem 29PQ

The amount of heat transferred in order for the whole process to take place is $\underset{\_}{2.83\times {10}^4\text{J}}$.

Explanation of Solution

Write the equation for the total heat transferred.

  $Q=Q_1+Q_2$                                                                                                                (I)

Here, $Q$ is the total heat energy, $Q_1$ is the heat energy required to raise the temperature to the melting point and $Q_2$ is the heat energy required to raise the temperature to the final temperature.

Write the expression of the heat energy required to raise the temperature to the melting point.

  $Q_1=mc\left(T_2-T_1\right)$                                                                                                         (II)

Here, $m$ is the mass, $c$ is the specific heat, $T_1$ is the initial temperature and $T_2$ is the final temperature.

Write the expression of the heat energy required to raise the temperature to final temperature.

  $Q_2=m{L}_f$                                                                                                                   (III)

Here, $L_f$ is the latent heat.

Rewrite the expression for the total heat transferred from equation (I) by using (II) and (III).

  $Q=mc\left(T_2-T_1\right)+m{L}_f$                                                                                             (IV)

Conclusion:

Substitute $45.75\text{g}$ for $m$, $386\text{J/kg-K}$ for $c$, $1361\text{K}$ for $T_2$, $22.3\text{°C}$ for $T_1$ and $2.07\times {10}^5\text{J/kg}$ for $L_f$ in equation (IV) to find $Q$.

  $\begin{array}{c}Q=\left[\left\{\left(45.75\text{g}\right)\left(\frac{1\times {10}^{-3}\text{kg}}{1\text{g}}\right)\right\}\left(386\text{J/kg-K}\right)\left(\left(1361\text{K}\right)-\left(22.3°\text{C+273}\right)\right)\right]+\\ \left[\left\{\left(45.75\text{g}\right)\left(\frac{1\times {10}^{-3}\text{kg}}{1\text{g}}\right)\right\}\left(2.07\times {10}^5\text{J/kg}\right)\right]\\ =\left(18800\text{J}\right)+\left(9470\text{J}\right)\\ =2.83\times {10}^4\text{J}\end{array}$

Thus, the amount of heat transferred in order for the whole process to take place is $\underset{\_}{2.83\times {10}^4\text{J}}$.

(b)

To determine

The amount of heat transferred to raise copper to its melting point.

Answer to Problem 29PQ

The amount of heat transferred to raise copper to its melting point is $\underset{\_}{1.88\times {10}^4\text{J}}$.

Explanation of Solution

Write the expression of the heat energy required to raise the temperature to the melting point.

  $Q_1=mc\left(T_2-T_1\right)$                                                                                                         (V)

Here, $Q_1$ is the heat energy required to raise the temperature to the melting point, $m$ is the mass, $c$ is the specific heat, $T_1$ is the initial temperature and $T_2$ is the final temperature.

Conclusion:

Substitute, $45.75\text{g}$ for $m$, $386\text{J/kg-K}$ for $c$, $1361\text{K}$ for $T_2$, $22.3\text{°C}$ for $T_1$ in equation (V) to find $Q_1$.

  $\begin{array}{c}{Q}_1=\left[\left\{\left(45.75\text{g}\right)\left(\frac{1\times {10}^{-3}\text{kg}}{1\text{g}}\right)\right\}\left(386\text{J/kg-K}\right)\left(\left(1361\text{K}\right)-\left(22.3°\text{C+273}\right)\right)\right]\\ =\left[\left\{\left(45.75\text{g}\right)\left(\frac{1\times {10}^{-3}\text{kg}}{1\text{g}}\right)\right\}\left(386\text{J/kg-K}\right)\left(\left(1361\text{K}\right)-\left(295\text{K}\right)\right)\right]\\ =\left(1.88\times {10}^4\text{J}\right)\end{array}$

Thus, the amount of heat transferred to raise copper to its melting point is $\underset{\_}{1.88\times {10}^4\text{J}}$.

(c)

To determine

The amount of heat transferred during phase change.

Answer to Problem 29PQ

The amount of heat transferred during phase change is $\underset{\_}{9.47\times {10}^3\text{J}}$.

Explanation of Solution

Write the expression of the heat energy required phase change.

  $Q_2=m{L}_f$                                                                                                                   (VI)

Here, $Q_2$ is the heat energy required to phase change, $L_f$ is the latent heat.

Conclusion:

Substitute $45.75\text{g}$ for $m$ and $2.07\times {10}^5\text{J/kg}$ for $L_f$ in equation (VI) to find $Q_2$.

  $\begin{array}{c}{Q}_2=\left[\left\{\left(45.75\text{g}\right)\left(\frac{1\times {10}^{-3}\text{kg}}{1\text{g}}\right)\right\}\left(2.07\times {10}^5\text{J/kg}\right)\right]\\ =\left[\left(45.75\times {10}^{-3}\text{kg}\right)\left(2.07\times {10}^5\text{J/kg}\right)\right]\\ =\left(9.470\times {10}^3\text{J}\right)\end{array}$

Thus, the amount of heat transferred during phase change is $\underset{\_}{9.47\times {10}^3\text{J}}$.

The approximation is the specific heat of solid copper and the liquid copper is considered as same. The energy is required for copper is about double comparing to the silver.