Unit 9 · Topic 9.5

9.5 Specific Heat and Thermal Conductivity

Specific heat tells you how much energy it takes to change a material's temperature, through Q = mcΔT. Thermal conductivity tells you how fast energy flows through a material by conduction, and the rate grows with area and temperature difference and shrinks with thickness. Both depend on what the material is made of.

Key terms

specific heat
thermal conductivity
rate of energy transfer
temperature difference
thickness and cross-sectional area

Specific heat

Different materials need different amounts of energy to warm up by the same amount. The specific heat c is the energy needed to raise 1 kg of a material by 1 K (or 1 °C, which is the same size step). The energy transferred is:

$Q = mc\Delta T$

Water has a very large specific heat, about 4190 J/(kg·K). Aluminum's is about 900 J/(kg·K), and iron's about 450 J/(kg·K). That's why a pot of water takes so long to boil, and why coastal towns have milder weather: the ocean soaks up and releases huge amounts of energy with small temperature changes.

Specific heat is a property of the material itself, set by how its atoms are arranged and how they interact. In this course it's treated as constant, not changing with temperature.

Mixing problems

When a hot object and a cold one share energy inside an insulated container, the energy one loses equals the energy the other gains. Write it as the total of all the Q's equal to zero:

$m_1c_1(T_f - T_1) + m_2c_2(T_f - T_2) = 0$

Using ( $T_f$ − $T_{\text{initial}}$ ) for both objects keeps the signs automatic: the hot object's Q comes out negative and the cold object's positive.

Thermal conductivity

Thermal conductivity k measures how well a material conducts energy. The rate of energy flow through a slab of material is:

$\frac{Q}{\Delta t} = \frac{kA\Delta T}{L}$

A is the cross-sectional area the energy passes through, L is the thickness, and ΔT is the temperature difference between the two faces. The rate is in watts (J/s).

Double the area: twice the rate (more paths for the energy).
Double the temperature difference: twice the rate.
Double the thickness: half the rate.
Metals have large k; air, foam and wool have small k. Insulation works by trapping air.

In the lab

To measure a specific heat, you might warm a known mass with an electric heater of known power, so Q = PΔt, and record the temperature over time. A graph of temperature against energy added is a straight line with slope $\dfrac{1}{mc}$ , so c = 1/(m × slope). Energy lost to the surroundings makes the measured c come out too large, since some of the energy you counted didn't warm the sample.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Warming water
How much energy does it take to heat 0.50 kg of water from 20 °C to 80 °C? Use c = 4190 J/(kg·K).
Show the solutionHide the solution
1. Step 1: ΔT = 80 − 20 = 60 °C, which is also 60 K.
2. Step 2: Q = mcΔT = (0.50)(4190)(60) = 125,700 J.
Answer: About 1.3 × 10⁵ J
Example 2Calculator allowed
Hot metal in cool water
A 0.20 kg aluminum block (c = 900 J/(kg·K)) at 90 °C is dropped into 0.50 kg of water (c = 4190 J/(kg·K)) at 20 °C in an insulated cup. Find the final temperature.
Show the solutionHide the solution
1. Step 1: Energy lost by the aluminum = energy gained by the water: $(0.20)(900)(T_f - 90) + (0.50)(4190)(T_f - 20) = 0$ .
2. Step 2: Expand: $180T_f$ − 16,200 + $2095T_f$ − 41,900 = 0, so $2275T_f$ = 58,100.
3. Step 3: $T_f$ ≈ 25.5 °C.
4. Step 4: Check: the aluminum loses (0.20)(900)(64.5) ≈ 11,600 J and the water gains (0.50)(4190)(5.5) ≈ 11,600 J. The final temperature is close to the water's, because water has the larger mass and far larger specific heat.
Answer: About 25.5 °C
Example 3Calculator allowed
Insulating a cooler
A foam cooler wall has k = 0.030 W/(m·K), area 0.80 m² and thickness 2.0 cm. Outside is 30 °C and inside is 5 °C. At what rate does energy conduct in? If you doubled the wall thickness, what would the rate be?
Show the solutionHide the solution
1. Step 1: Convert thickness: L = 0.020 m. ΔT = 30 − 5 = 25 K.
2. Step 2: $\dfrac{Q}{\Delta t} = \dfrac{kA\Delta T}{L} = \dfrac{(0.030)(0.80)(25)}{0.020} = 30$ W.
3. Step 3: The rate is inversely proportional to L, so doubling the thickness halves it to 15 W.
Answer: 30 W; doubling the thickness gives 15 W

Common mistakes

Forgetting to convert thickness to meters in the conduction equation. Centimeters make the rate 100 times too small.
Assuming two objects in contact change temperature by the same amount. The one with the smaller mc changes more.
Mixing up specific heat (how much energy to change temperature) with thermal conductivity (how fast energy moves through).

On the exam

Lab questions may ask you to design a procedure to find a specific heat or conductivity, choose what to graph, and use the slope.
Factor-of-change questions on the conduction equation are common: change A, L or ΔT and say what happens to the rate.

Connected topics

Videos

AP Physics 2 - Unit 9 - Lesson 5 - Heat Capacity
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Specific heat capacity | Khan Academy
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AP Physics 2 - Unit 9 - Lesson 6 - Thermal Conductivity
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Check yourself

4 questions on 9.5 Specific Heat and Thermal Conductivity. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

How much energy is needed to heat 0.50 kg of water from 20 °C to 80 °C? The specific heat of water is $4.2 \times 10^3$ J/(kg·K).

Question 2 of 4Calculator allowed

Equal masses of aluminum (specific heat 900 J/(kg·K)) and copper (specific heat 385 J/(kg·K)) each absorb the same amount of energy. How do their temperature changes compare?

Question 3 of 4Calculator allowed

A glass window pane has an area of 1.5 m² and a thickness of 4.0 mm. The glass has a thermal conductivity of 0.80 W/(m·K). The inside surface is 15 K warmer than the outside surface. At what rate is energy conducted through the glass?

Question 4 of 4Calculator allowed

Energy is conducted through a slab at rate H. A second slab of the same material has twice the area and twice the thickness, and the temperature difference across it is three times as large. At what rate is energy conducted through the second slab?

0 of 4 answered