Specific Heat
1. Heating a Block of Copper.
Follow the link: Example 22.1.
2. (Calculus) Low Temperature Study of Aluminum.
Follow the link: Example 22.2.
3. Low Temperature Specific Heat Varying with Temperature.
Follow the link: Exercise 22.1.5.1.
Section 22.4 Thermal Properties Bootcamp
Exercises Exercises
States of Matter
4. Heat Needed to Convert Ice to Water Vapor at High Temperature.
Follow the link: Example 22.7.
Calorimetry
5. Calorimetry of Mixing Water at Different Temperatures.
Follow the link: Example 22.9.
6. Calorimetry of Adding Warm Water to Low Temperature Ice.
Follow the link: Exercise 22.3.1.
7. Calorimetry of Melting of Ice in a Brass Container.
Follow the link: Exercise 22.3.3.
8. Calorimetry of Adding Warm Water to Low Temperature Ice.
Follow the link: Exercise 22.3.1.
9. Calorimetry of Mixing Ice and Steam.
Follow the link: Exercise 22.3.2.
Miscellaneous
10. Thermodynamics of Aerobics.
During an aerobic exercise a person of mass \(60\, \text{kg}\) produces 250 kcal energy per hour from her metabolic rate. Suppose \(25\%\) of this energy goes into aerobic workout and the rest is converted into heat which is used to evaporate the sweat. Assume the heat of evaporation at the body temperature of \(37^{\circ}\text{C}\) to be approximately \(2.4 \times 10^6\, \text{J/kg}\text{.}\) What is the rate at which water is lost from the body?
Solution.
Let us first convert the unit of energy in the workout to \(J\text{.}\) The rate at which the energy is spent in the workout is given to be
\begin{align*}
P \amp = \textrm{250 kcal energy per hour}\\
\amp = \frac{250\ \textrm{kcal}}{\textrm{h}} \times \frac{4180\ \textrm{J}}{1\ \textrm{kcal}} \frac{1\ \textrm{h} }{3600\ \textrm{s}} = 290.3 \textrm{J/s}.
\end{align*}
The problem states that \(75\%\) of this energy goes towards evaporation of the sweat. Let \(m\) be the mass evaporated each second.
\begin{align*}
m \amp = \frac{0.75\times 290.3 \textrm{J}}{ l_{\textrm{evap}}} = \frac{0.75\times 290.3 \textrm{J}}{2.4 \times 10^6\ \textrm{J/kg}} = 9.07\times 10^{-5}\ \textrm{kg}.
\end{align*}
This says that only a very minute amount of the water is evaporated during the workout.
11. Heating a Copper Block by Passing a Current.
A wire of resistance \(2\) Ohms is wound around a copper cylinder of mass \(200\) g. The arrangement is insulated so that all heat generated when a current is passed in the wire goes to raise the temperature of the wire and copper. Now, you pass \(5\)-A current through the wire for 10 sec. As a result, the temperature of copper cylinder rises by \(6.5^{\circ}\text{C}\text{.}\) The experiment is done at atmospheric pressure.
(a) What value of specific heat of copper does this experiment yield? (b) What would be molar specific heat of copper based on the data of this experiment? (c) What is the heat capacity of the copper cylinder? Ignore the effect of the heating of the current-carrying wire itself. Use: Energy given to the wire of resistance \(R\) by current \(I\) over time \(\Delta t\) equals \(I^2 R\Delta t\text{.}\)
Solution 1. a
The energy produced by passing the current is given by
\begin{equation*}
E = I^2 R \Delta t = (5\ \textrm{A})^2 \times (2\ \Omega)\times 10\ \textrm{s} = 500\ \textrm{J}.
\end{equation*}
This energy will go towards raising the temperature of the wire and the copper cylinder. Assuming negligible mass for the wires themselves we ignore the energy into the wires. This gives
\begin{equation*}
m_{Cu} c \Delta T = 500\ \textrm{J}.
\end{equation*}
Given the mass of the cylinder and the change in the temperature, the specific heat of copper is found to be
\begin{equation*}
c = \frac{500\ \textrm{J}}{m_{Cu} \Delta T } = \frac{500\ \textrm{J}}{0.2\ \textrm{kg} \times 6.5^{\circ}\textrm{C} } = 385\ \frac{\textrm{J}}{ \textrm{kg.} ^{\circ}\textrm{C}}.
\end{equation*}
Solution 2. b
The molecular weight of copper is 0.063 kg/mol. Therefore, the molar specific heat
\begin{equation*}
C = c \times \textrm{mol. wt.} = 385\ \frac{\textrm{J}}{ \textrm{kg.} ^{\circ}\textrm{C}} \times 0.063 \textrm{kg/mol} = 24.3\ \frac{\textrm{J}}{ \textrm{mol.} ^{\circ}\textrm{C}}
\end{equation*}
Solution 3. c
The heat capacity is equal to the product of the specific heat and the mass of the system.
\begin{equation*}
\textrm{Heat capacity} = 385\ \frac{\textrm{J}}{ \textrm{kg.} ^{\circ}\textrm{C}} \times 0.2 \ \textrm{kg} = 77 \ \textrm{J}/^{\circ}\textrm{C}.
\end{equation*}
