Specific heat and thermal conductivity

Energy Required to Change Temperature

Every material needs a different amount of energy to warm up by the same amount. Specific heat tells you exactly how much energy one kilogram of a particular material requires per degree of temperature change.

Key Equations

Energy transferred during temperature change:

$$Q=mc\Delta T$$

Variables:

• $Q$ = energy transferred (J)
• $m$ = mass (kg)
• $c$ = specific heat of the material (J/(kg·°C) or J/(kg·K))
• $\Delta T$ = change in temperature (°C or K)

Rearrangements:

$$c=\frac{Q}{m\Delta T}$$

$$\Delta T=\frac{Q}{mc}$$

$$m=\frac{Q}{c\Delta T}$$

ProportionalityThe energy transferred is directly proportional to mass, directly proportional to specific heat, and directly proportional to the temperature change. Doubling the mass doubles the energy required for the same temperature change.

Reference sheet status: On the reference sheet.

The symbol $\Delta T$ means the final temperature minus the initial temperature: $\Delta T={T}_{f}-{T}_{i}$. A positive $\Delta T$ means the material warmed up. A negative $\Delta T$ means it cooled down.

The specific heat $c$ is an intrinsic property of a material. It depends on how atoms are arranged within the material and how those atoms interact with each other. Metals like copper have low specific heats, so they warm quickly. Water has a high specific heat, so it resists temperature changes.

Because specific heat is intrinsic, it does not depend on the size or shape of the sample. A 1 kg block of copper and a 50 kg block of copper share the same specific heat. However, the larger block needs more total energy because $Q$ is proportional to $m$.

When two objects at different temperatures are placed in thermal contact, energy transfers from the hotter object to the cooler one. If the system is thermally isolated, the energy lost by the hot object equals the energy gained by the cool object. This means ${Q}_{\text{lost}}+{Q}_{\text{gained}}=0$, because energy is conserved within the system.

BoundaryAP Physics 2 models specific heat as independent of temperature. Real materials have specific heats that vary with temperature, but this is outside the scope of the AP exam.

MisconceptionStudents often assume that two objects receiving the same energy will reach the same final temperature. Objects with different masses or specific heats will have different $\Delta T$ values even for the same $Q$.

Exam TipAlways check whether $m$ and $c$ are the same before assuming equal temperature changes.

Rate of Energy Transfer by Conduction

Thermal conduction is the transfer of energy through a material without the material itself moving. Energy passes from atom to atom through vibrations and collisions.

Key Equations

Rate of energy transfer by conduction:

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

Variables:

• $\frac{Q}{\Delta t}$ = rate of energy transfer, or thermal power (W)
• $k$ = thermal conductivity of the material (W/(m·K))
• $A$ = cross-sectional area perpendicular to the energy flow (m²)
• $\Delta T$ = temperature difference across the material (K or °C)
• $L$ = thickness of the material along the direction of energy flow (m)

Rearrangements:

$$k=\frac{QL}{\Delta t\cdot A\cdot \Delta T}$$

$$L=\frac{kA\Delta T}{\left(\frac{Q}{\Delta t}\right)}$$

ProportionalityThe rate of energy transfer is directly proportional to thermal conductivity, directly proportional to cross-sectional area, directly proportional to the temperature difference, and inversely proportional to thickness. Doubling the thickness halves the rate of energy transfer.

Reference sheet status: On the reference sheet.

In this equation, $\frac{Q}{\Delta t}$ has units of watts (W), because it represents energy transferred per unit time. The symbol $\Delta T$ here is the temperature difference between the two faces of the material, not a temperature change over time.

Thermal conductivity $k$ is an intrinsic property of the material. It depends on the arrangement and interactions of the atoms within the material. Metals have high thermal conductivities because free electrons transfer energy rapidly. Insulators like wood or foam have low thermal conductivities because energy transfers more slowly through atomic vibrations alone.

Because $k$ is intrinsic, it does not change when you alter the shape or size of the sample. A thin copper sheet and a thick copper rod share the same $k$ value. The rate of energy transfer differs because $A$ and $L$ differ, but the material property is the same.

The physical dimensions of the material matter in a clear way. A larger cross-sectional area $A$ provides more pathways for energy to flow, increasing the rate. A greater thickness $L$ means atoms must pass energy through more “layers,” decreasing the rate.

Examiner InsightAP questions often change one variable (area, thickness, or $\Delta T$) and ask how the rate changes. Set up a ratio of the equation before and after the change, and cancel unchanged quantities.

Exam TipWrite $\frac{(Q/\Delta t{)}_{\text{new}}}{(Q/\Delta t{)}_{\text{old}}}$ and simplify.