Lesson 9.7 · Unit 9 · CED topic 9.5
Specific heat and thermal conductivity
On a summer afternoon the sand at the beach is too hot to walk on, while the water beside it is pleasantly cool, even though both have been in the same sunshine for hours. Later, at night, the sand is cold and the water is still warm. The sand changes temperature easily and the water resists change. A thick wool blanket keeps you warm, while a thin sheet of metal does not, though both are the same temperature as the room. Two properties of a material explain these everyday facts: its specific heat, which says how much energy it takes to change its temperature, and its thermal conductivity, which says how fast energy moves through it. This lesson develops both, and shows how to measure them.
Describe the energy required to change the temperature of an object by a certain amount.
- The amount of energy required to change the temperature of a material is related to the material’s specific heat: \(Q = mc\Delta T\).
- The specific heat of a material is an intrinsic property of that material that depends on the arrangement and interactions of the atoms that make up the material.
Describe the rate at which energy is transferred by conduction through a given material.
- The rate at which energy is transferred by conduction through a given material is related to the thermal conductivity, the physical dimensions of the material, and the temperature difference across the material: \(\dfrac{Q}{\Delta t} = \dfrac{kA\Delta T}{L}\).
- The thermal conductivity of a material is an intrinsic property of that material that depends on the arrangement and interactions of the atoms that make up the material.
- Boundary: AP Physics 2 models specific heat as independent of temperature.
Specific heat
The energy needed to raise the temperature of a sample by \(\Delta T\) is proportional to its mass and to the temperature change: \[ Q = mc\,\Delta T, \] where \(c\) is the specific heat of the material, in J/(kg·K). Water has \(c = 4186\) J/(kg·K), which is large; copper has 385 and aluminum 900. A kelvin and a Celsius degree are the same size, so \(\Delta T\) can be taken in either. A larger \(c\) means more energy for the same temperature change, so a substance with a large specific heat warms and cools slowly. That is why water moderates coastal climates and why a car radiator uses water.
The specific heat is a property of the material’s atoms, not of the object’s size, and it depends on how the atoms are arranged and how they are bound. Water is unusual because its molecules can store energy in rotation and in hydrogen-bond stretching as well as in translation, so it takes a lot of energy to raise its temperature. A monatomic gas, by contrast, can store energy only as translational kinetic energy, so its internal energy increases by exactly \(\tfrac{3}{2}nR\) per kelvin.
When two objects exchange energy only with each other, the energy lost by one equals the energy gained by the other, \(Q_{hot} + Q_{cold} = 0\). The result is the calorimetry method: write \(mc\Delta T\) for each object (with \(\Delta T = T_f - T_i\), so a loss comes out negative), set the sum equal to zero, and solve for the common final temperature. With two objects, \(T_f\) is a weighted average of the starting temperatures, weighted by \(mc\), the energy each object stores per kelvin.
An experiment that measures \(c\): heat a sample with an electric heater of known power \(P\) for a time \(t\), so \(Q = Pt\), while recording the temperature. A graph of \(\Delta T\) against \(Q\) is a straight line with slope \(1/(mc)\); a graph of \(Q\) against \(\Delta T\) has slope \(mc\). Good practice includes insulating the sample (to reduce losses by conduction, convection and radiation), stirring liquids so the temperature is uniform, and subtracting the energy absorbed by the container.
Thermal conductivity
How quickly does energy move through a slab of material between a hot side and a cold side? Experiments show that the rate of energy transfer by conduction is proportional to the area \(A\) through which it passes, proportional to the temperature difference \(\Delta T\) across the slab, and inversely proportional to the thickness \(L\): \[ \frac{Q}{\Delta t} = \frac{kA\Delta T}{L}. \] The proportionality constant \(k\) is the thermal conductivity, in W/(m·K). A large \(k\) means a good conductor (copper, 400; aluminum, 240; window glass, 0.8; wood, 0.12; still air, 0.025). Think of it as a flow through a pipe: a wider pipe (larger area) and a bigger push (larger \(\Delta T\)) increase the flow; a longer pipe (larger \(L\)) decreases it.
- Doubling the thickness halves the rate.
- Doubling the area doubles the rate.
- Doubling the temperature difference doubles the rate.
- Replacing glass with foam (smaller \(k\)) cuts the rate by the ratio of the conductivities.
The conductivity depends on the atomic structure, which is why metals (with free electrons that carry energy) are good conductors and gases (with widely spaced atoms) are poor ones. A good insulator, such as a double-pane window with an air gap, uses a layer of low-\(k\) material to reduce the rate. Experiments on conductivity usually measure the temperatures on both sides of a slab of known size and the energy that passes in a known time, or the rate at which an ice-water bath melts. They often linearize by plotting the rate against \(A\Delta T/L\); the slope is \(k\).
How much energy is required to heat 0.50 kg of water from 20\(^\circ\)C to 50\(^\circ\)C? How long does a 1000 W heater (no losses) take?
Equation. \(Q = mc\Delta T = (0.50)(4186)(30) = 6.28\times10^4\) J.
Time. \(t = Q/P = 6.28\times10^4/1000 = 63\) s.
A 0.50 kg block of copper at 200\(^\circ\)C is dropped into 1.0 kg of water at 20\(^\circ\)C in an insulated container. Find the final temperature. (\(c_{Cu} = 385\), \(c_w = 4186\) J/(kg·K).)
Energy balance. \(m_{Cu}c_{Cu}(T_f - 200) + m_wc_w(T_f - 20) = 0\).
Solve. \((192.5)(T_f - 200) + (4186)(T_f - 20) = 0\), so \(4378.5\,T_f = 38500 + 83720\) and \(T_f = 27.9\,^\circ\)C.
Check. The water has much more energy capacity per kelvin than the copper (4186 against 192.5 J/K), so the final temperature is near the water’s start. ✓
A brick wall (\(k = 0.80\) W/(m·K)) is 0.20 m thick and 10 m\(^2\) in area. The inside is at 20\(^\circ\)C and the outside at 0\(^\circ\)C. Find the rate of energy transfer, and the energy lost in one hour.
Rate. \(Q/\Delta t = kA\Delta T/L = (0.80)(10)(20)/0.20 = 800\) W.
One hour. \(Q = (800)(3600) = 2.9\times10^6\) J.
Effect of insulation. Adding a 0.10 m layer of foam (\(k = 0.04\)) in series would reduce the rate a great deal, because most of the temperature drop would then occur across the foam.
- Specific heat. \(Q = mc\Delta T\); \(Q\) negative when the object cools. A larger \(c\) means the temperature changes less for the same energy.
- Calorimetry. Energy lost equals energy gained; final temperature is a weighted average.
- Conduction rate. \(kA\Delta T/L\): proportional to area and temperature difference; inversely proportional to thickness.
- Material properties. \(c\) and \(k\) are intrinsic; they depend on the atomic structure, not on the size of the sample.
- Experiments. Heater of known power: \(Q = Pt\). Linearize: \(Q\) against \(\Delta T\) (slope \(mc\)).
- How much energy is needed to raise 2.0 kg of aluminum (\(c = 900\) J/(kg·K)) by 40 K?
Show answer
\(Q = (2.0)(900)(40) = 7.2\times10^4\) J.
- The same energy is given to equal masses of water and copper, both starting at the same temperature. Which ends up hotter, and why?
Show answer
The copper, because it has the smaller specific heat (385 against 4186 J/(kg·K)), so the same \(Q\) gives a much larger \(\Delta T\).
- 0.20 kg of water at 80\(^\circ\)C is mixed with 0.20 kg of water at 20\(^\circ\)C in an insulated cup. Find the final temperature.
Show answer
Same material and equal masses: \(T_f = (80+20)/2 = 50\,^\circ\)C.
- The rate of energy transfer through a window is 120 W. If the glass is replaced by glass twice as thick of the same area and the same temperature difference, what is the new rate?
Show answer
The rate is inversely proportional to the thickness, so it becomes 60 W.
- A student plots \(Q\) against \(\Delta T\) for a 0.25 kg sample and finds a slope of 225 J/K. What is the specific heat?
Show answer
Slope \(= mc\), so \(c = 225/0.25 = 900\) J/(kg·K), consistent with aluminum.
- Explain why a coastal city has smaller day-to-night temperature swings than a desert city at the same latitude.
Show answer
Water has a large specific heat, so large amounts of energy must enter or leave it to change its temperature; the ocean warms little by day and cools little by night, and the air above it follows. The dry ground and air of the desert change temperature easily.