1.5 - Specific Heat and Thermal Conductivity

Specific Heat

How much energy it takes to change a material’s temperature depends on the material’s specific heat.

Q = mc\Delta T

where: -

Q

= energy transferred by heating (J) -

m

= mass (kg) -

c

= specific heat capacity (J/(kg·K) or J/(kg·°C)) -

\Delta T

= temperature change (K or °C)

Specific heat,

c

, is the energy needed to raise the temperature of 1 kg of a material by 1 K.

It’s an Intrinsic Property

A material’s specific heat is an intrinsic property, meaning it depends on how the atoms in the material are arranged and how they interact. Some values to compare:

Water: $c = 4186$ J/(kg·K), which is very high

Aluminum: $c = 900$ J/(kg·K)

Iron: $c = 450$ J/(kg·K)

Copper: $c = 385$ J/(kg·K)

A high specific heat means the material needs more energy to change its temperature, so it resists temperature change.

Why materials differ. Energy you add to a material is shared between the kinetic energy of its particles (which sets the temperature) and the potential energy of their interactions (bonds stretching and squeezing). How it's shared, and how many particles share it, depends on what the material is made of and how its atoms are arranged and bonded:

Water has a very high $c$ because a lot of the added energy goes into the hydrogen bonds between water molecules (potential energy) instead of into faster motion, so the temperature rises less.

Aluminum vs. copper: per atom, these two metals need about the same energy to warm up by 1 K. But aluminum atoms are much lighter, so a kilogram of aluminum has more than twice as many atoms to share the energy. That's why $c$ for aluminum (900 J/(kg·K)) is more than twice $c$ for copper (385 J/(kg·K)).

Intrinsic also means

c

doesn't depend on how much material you have. A 1 kg block and a 5 kg block of copper have the same

c

. The 5 kg block needs 5 times the energy for the same temperature change because

m

is 5 times bigger, not because

c

changed.

One note for this course: AP Physics 2 treats specific heat as constant, meaning it doesn’t depend on temperature.

Worked Example: Measuring Specific Heat

An electric heater delivers energy at a steady 50.0 W to a 0.500 kg block of an unknown metal. The block is wrapped in insulation, and its temperature is recorded every minute:

$t$ (s)	0	60	120	180	240
$T$ (°C)	20.0	26.7	33.3	40.0	46.7

Step 1: set up a straight line. After time

t

the heater has added

Q = P_{\text{heater}}t

, and

Q = mc\Delta T

, so

\Delta T = \left(\frac{P_{\text{heater}}}{mc}\right)t

A graph of temperature vs. time is a straight line with slope

P_{\text{heater}}/(mc)

Step 2: find the slope. The data fall on a straight line, so use its endpoints:

\text{slope} = \frac{46.7\,^{\circ}\text{C} - 20.0\,^{\circ}\text{C}}{240 \text{ s} - 0 \text{ s}} = 0.111 \text{ K/s}

Step 3: solve for c.

c = \frac{P_{\text{heater}}}{m(\text{slope})} = \frac{50.0 \text{ W}}{(0.500 \text{ kg})(0.111 \text{ K/s})} \approx 9.0 \times 10^2 \text{ J/(kg·K)}

That matches aluminum (900 J/(kg·K)), so the block is probably aluminum.

Which way does the error go? If some energy leaks out to the room, the block warms more slowly, the slope comes out too small, and the calculated

c

comes out too big. Insulating the block and keeping the run short keeps this error small.

Thermal Conductivity

How fast energy conducts through a material depends on the material’s thermal conductivity, its physical dimensions, and the temperature difference across it.

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

or, written as power:

P = \frac{kA\Delta T}{L}

where: -

\frac{Q}{\Delta t}

= rate of heat transfer, which is power (W) -

k

= thermal conductivity (W/(m·K)) -

A

= cross-sectional area (m²) -

\Delta T

= temperature difference (K) -

L

= length or thickness (m)

Also an Intrinsic Property

Thermal conductivity is another intrinsic property that depends on atomic arrangement and interactions.

Metals conduct well because they have free electrons that move through the whole material and carry energy quickly from hot regions to cold ones.

Nonmetal solids (wood, plastic, glass) have no free electrons, so energy passes only from each atom to its neighbors through their vibrations, which is much slower.

Gases like air conduct poorly because their molecules are far apart and collide less often. Trapped air is why foam, fiberglass, and down jackets insulate so well.

Good conductors have a high

k

: - Metals like copper, aluminum, and silver. - Roughly

k \sim 100

400

W/(m·K).

Poor conductors, or insulators, have a low

k

: - Wood, plastic, air. - Roughly

k \sim 0.02

0.2

W/(m·K).

How the rate depends on the size and shape:

Double the area $A$ : twice as many paths for energy, so the rate doubles.

Double the thickness $L$ : the same temperature difference is spread over twice the distance, so the rate is cut in half.

Double the temperature difference $\Delta T$ : the rate doubles.

Example (factor of change): a wall is rebuilt 50% thicker, and its area is doubled. Same material, same indoor and outdoor temperatures. The new rate compared with the old one is

\frac{P_{\text{new}}}{P_{\text{old}}} = \frac{k(2A)\Delta T/(1.5L)}{kA\Delta T/L} = \frac{2}{1.5} = 1.33

so energy leaks through 1.33 times as fast.

Example (numbers): a copper rod (

k = 4.0 \times 10^2

W/(m·K)) is 0.50 m long with a cross-sectional area of

2.0 \times 10^{-4}

m² (2.0 cm²). One end sits in boiling water at 100°C and the other in ice water at 0°C:

\frac{Q}{\Delta t} = \frac{kA\Delta T}{L} = \frac{(4.0 \times 10^2 \text{ W/(m·K)})(2.0 \times 10^{-4} \text{ m}^2)(100 \text{ K})}{0.50 \text{ m}} = 16 \text{ W}

In 5.0 minutes (300 s), the rod carries

Q = (16 \text{ W})(300 \text{ s}) = 4.8 \times 10^3

J into the ice water.

Thermal Resistance

You can rewrite the conduction equation as:

P = \frac{\Delta T}{R_{\text{thermal}}}

where the thermal resistance is

R_{\text{thermal}} = \frac{L}{kA}

This has a nice analogy to electricity, where

I = \frac{\Delta V}{R}

: - Heat flow is like current. - Temperature difference is like voltage. - Thermal resistance is like electrical resistance.

Practice Problems

1. How much energy does it take to raise the temperature of 2.0 kg of water from 20°C to 80°C? (

c_{\text{water}} = 4186

J/(kg·K))

2. A 0.50 kg aluminum block (

c = 900

J/(kg·K)) at 100°C is dropped into 2.0 kg of water at 20°C. What is the final equilibrium temperature?

3. A glass window has an area of 2.0 m², a thickness of 5.0 mm, and a thermal conductivity of 0.8 W/(m·K). If it’s 20°C inside and 0°C outside, what is the rate of heat loss?

4. Two materials have the same cross-sectional area and length. Material A has twice the thermal conductivity of material B. Compare their heat transfer rates for the same temperature difference.

5. Why does metal feel colder than wood when they’re both at the same temperature?

Solutions

Q = mc\Delta T = (2.0)(4186)(80-20) = 502,000 \text{ J} = 502 \text{ kJ}

The energy lost by the aluminum equals the energy gained by the water:

m_{\text{Al}}c_{\text{Al}}(100-T_f) = m_{\text{w}}c_{\text{w}}(T_f-20)

(0.5)(900)(100-T_f) = (2.0)(4186)(T_f-20)

45,000 - 450T_f = 8372T_f - 167,440

212,440 = 8822T_f

T_f = 24.1°C

Convert first:

L = 5.0

= 5.0 \times 10^{-3}

m, and

\Delta T = 20

P = \frac{kA\Delta T}{L} = \frac{(0.8)(2.0)(20)}{5.0 \times 10^{-3}} = 6400 \text{ W} = 6.4 \text{ kW}

Since

P \propto k

when everything else is equal:

\frac{P_A}{P_B} = \frac{k_A}{k_B} = \frac{2k_B}{k_B} = 2

Material A moves heat twice as fast.

5. Metal has a much higher thermal conductivity than wood.

When you touch metal: - Heat conducts out of your hand into the metal quickly. - You lose a lot of heat from your skin. - So it feels cold.

When you touch wood: - Heat conducts out of your hand slowly. - You lose only a little heat from your skin. - So it feels warmer.

They’re both at the same temperature, but the metal pulls heat away from your hand faster, which is what makes it feel cold.

Common Mistakes

Wrong temperature units. $\Delta T$ can be in K or °C since the size of the change is the same, but $c$ has to use consistent units.

Confusing Q and Q/Δt. $Q$ is total energy (J), while $\frac{Q}{\Delta t}$ is a rate, or power (W).

Forgetting mass in specific heat. The energy needed depends on both the specific heat and the mass.

Using diameter instead of area. The conduction equation uses cross-sectional area, not diameter.

Confusing thermal conductivity with specific heat. $k$ sets the rate of heat transfer, while $c$ sets the temperature change for a given amount of energy.

Using the wrong length in conduction. $L$ is the distance the heat travels, which is the thickness, not the perimeter or some other dimension.

Assuming the equilibrium temperature is just the average. You have to use conservation of energy. Heavier objects, or ones with higher specific heat, “win.”

Not converting units. Thickness is often given in mm or cm, so convert to meters.

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