Chapter 21, Problem 20PQ

From Table 21.1, the specific heat of milk is 3.93 × 10³ J/ (kg ∙ K). and the specific heat of water is 4.19 × 10³ J/(kg ∙ K). Suppose you wish to make a large mug (0.500 L) of hot chocolate. Each liquid is initially at 5.00°C. and you need to raise their temperature to 80.0°C. The density of milk is about 1.03 × 10³ kg/m³, and the density of water is 1.00 × 10³ kg/m³. a. How much heat must be transferred in each case? b. If you use a small electric hot plate that puts out 455 W, how long would it take to heat each liquid?

To determine

The amount of heat transferred to milk and water.

Answer to Problem 20PQ

The amount of heat transferred to milk is $Q_m=1.52\times {10}^5\text{J}$ and water is $Q_w=1.57\times {10}^5\text{J}$.

Explanation of Solution

Write the expression for heat energy transferred to the system.

  $Q=mc\Delta T$                                                                                                     (I)

Here, $Q$ is the heat energy, $m$ is the mass, $c$ is the specific heat capacity of the object, and $\Delta T$ is the change in temperature.

Write the expression for mass of the milk, relating density and volume.

  $m_m={\rho }_m{V}_m$                                                                                                    (II)

Here, $m_m$ is the mass of the milk, ${\rho }_m$ is the density of milk, and $V_m$ is the volume of the milk container.

Write the expression for mass of the water, relating density and volume.

  $m_w={\rho }_w{V}_w$                                                                                                    (III)

Here, $m_w$ is the mass of the water, ${\rho }_w$ is the density of water, and $V_w$ is the volume of the water container.

Rearrange the equation (I) to calculate the heat required for milk.

  $Q_m=m_m{c}_m\Delta T$                                                                                             (IV)

Here, $Q_m$ is the heat energy required for milk and $c_m$ is the specific heat capacity of milk.

Rearrange the equation (I) to calculate the heat required for water.

  $Q_w=m_w{c}_w\Delta T$                                                                                              (V)

Here, $Q_w$ is the heat energy required for water and $c_w$ is the specific heat capacity of water.

Conclusion:

Substitute $1.03\times {10}^3{\text{kg/m}}^3$ for ${\rho }_m$ and $0.500\text{L}$ for $V_m$ in equation (II) to find ${\rho }_m$.

  $\begin{array}{c}{m}_m=\left(1.03\times {10}^3{\text{kg/m}}^3\right)\left(0.500\text{L}\right)\left(\frac{0.001{\text{m}}^3}{1.0\text{L}}\right)\\ =0.515\text{kg}\end{array}$

Substitute $1.00\times {10}^3{\text{kg/m}}^3$ for ${\rho }_w$ and $0.500\text{L}$ for $V_w$ in equation (III) to find ${\rho }_w$.

  $\begin{array}{c}{m}_w=\left(1.00\times {10}^3{\text{kg/m}}^3\right)\left(0.500\text{L}\right)\left(\frac{0.001{\text{m}}^3}{1.0\text{L}}\right)\\ =0.500\text{kg}\end{array}$

Substitute $0.515\text{kg}$ for $m_m$, $3.93\times {10}^3\text{J/kg}\cdot \text{K}$ for $c_m$, and $80.0°\text{C}-5.00°\text{C}$ for $\Delta T$ in equation (IV) to find $Q_m$.

  $\begin{array}{c}{Q}_m=\left(0.515\text{kg}\right)\left(3.93\times {10}^3\text{J/kg}\cdot \text{K}\right)\left(80.0°\text{C}-5.00°\text{C}\right)\\ =\left(0.515\text{kg}\right)\left(3.93\times {10}^3\text{J/kg}\cdot \text{K}\right)\left(75.0°\text{C}\right)\\ =1.52\times {10}^5\text{J}\end{array}$

Substitute $0.500\text{kg}$ for $m_w$, $4.19\times {10}^3\text{J/kg}\cdot \text{K}$ for $c_w$, and $80.0°\text{C}-5.00°\text{C}$ for $\Delta T$ in equation (V) to find $Q_w$.

  $\begin{array}{c}{Q}_w=\left(0.500\text{kg}\right)\left(4.19\times {10}^3\text{J/kg}\cdot \text{K}\right)\left(80.0°\text{C}-5.00°\text{C}\right)\\ =\left(0.500\text{kg}\right)\left(4.19\times {10}^3\text{J/kg}\cdot \text{K}\right)\left(75.0°\text{C}\right)\\ =1.57\times {10}^5\text{J}\end{array}$

Therefore, the amount of heat transferred to milk is $Q_m=1.52\times {10}^5\text{J}$ and water is $Q_w=1.57\times {10}^5\text{J}$.

To determine

The time taken to heat milk and water.

Answer to Problem 20PQ

The time taken to heat milk is $3.34\times {10}^2\text{s}$ and water is $3.45\times {10}^2\text{s}$.

Explanation of Solution

Write the expression for power (energy per unit time).

  $P=\frac{Q}{\Delta t}$

Here, $P$ is the power of the electric hot plate and $\Delta t$ is the time taken to heat.

Rearrange the above equation for $\Delta t$.

  $\Delta t=\frac{Q}{P}$                                                                                                                  (VI)

Rearrange the equation for time taken to heat the milk.

  $\Delta t_m=\frac{Q_m}{P}$                                                                                                            (VII)

Here, $\Delta t_m$ is the time taken to heat the milk.

Rearrange the equation for time taken to heat the water.

  $\Delta t_w=\frac{Q_w}{P}$                                                                                                            (VIII)

Here, $\Delta t_w$ is the time taken to heat the water.

Conclusion:

Substitute $1.52\times {10}^5\text{J}$ for $Q_m$ and $455\text{W}$ for $P$ in equation (VII) to find $\Delta t_m$.

  $\begin{array}{c}\Delta t_m=\frac{1.52\times {10}^5\text{J}}{455\text{W}}\\ =3.34\times {10}^2\text{s}\end{array}$

Substitute $1.57\times {10}^5\text{J}$ for $Q_w$ and $455\text{W}$ for $P$ in equation (VIII) to find $\Delta t_w$.

  $\begin{array}{c}\Delta t_w=\frac{1.57\times {10}^5\text{J}}{455\text{W}}\\ =3.45\times {10}^2\text{s}\end{array}$

Therefore, the time taken to heat milk is $3.34\times {10}^2\text{s}$ and water is $3.45\times {10}^2\text{s}$.