AP Physics 2- 9.5 Specific Heat and Thermal Conductivity- Study Notes- New Syllabus
AP Physics 2- 9.5 Specific Heat and Thermal Conductivity – Study Notes
AP Physics 2- 9.5 Specific Heat and Thermal Conductivity – Study Notes – per latest Syllabus.
Key Concepts:
- Specific Heat
- Thermal Conductivity
Specific Heat
The energy required to change the temperature of an object is the heat energy absorbed or released by the object when its temperature changes without a phase change.
Key Concepts:
Heat Capacity (C): The amount of heat required to raise the temperature of an object by 1 K (or 1 °C).
\( Q = C \Delta T \)
Specific Heat Capacity (c): The heat required to raise the temperature of 1 kg of a substance by 1 K.
\( Q = m c \Delta T \)
- \( Q \): Heat added or removed (Joules)
- \( m \): Mass of the object (kg)
- \( \Delta T = T_{final} – T_{initial} \)
- Positive \( Q \): Heat absorbed → Temperature increases
- Negative \( Q \): Heat released → Temperature decreases
Relationship with Internal Energy:
For a system with constant volume:
\( Q = \Delta U = m c_v \Delta T \)
Where \( c_v \) is the specific heat at constant volume.
Example :
How much heat is required to raise the temperature of 0.5 kg of water from 25 °C to 75 °C? (Take \( c = 4200 \, J/kg·K \))
▶️ Answer/Explanation
\( Q = m c \Delta T \)
\( = 0.5 \times 4200 \times (75 – 25) \)
\( = 0.5 \times 4200 \times 50 = 105,000 \, J \)
Answer: Heat required = 105 kJ.
Example :
A 2 kg block of copper (\( c = 390 \, J/kg·K \)) is heated from 20 °C to 80 °C. Find the heat energy required.
▶️ Answer/Explanation
\( Q = m c \Delta T \)
\( = 2 \times 390 \times (80 – 20) \)
\( = 2 \times 390 \times 60 = 46,800 \, J \)
Answer: Heat required = 46.8 kJ.
Thermal Conductivity
Conduction is the transfer of heat through a material without bulk movement of the material. The rate at which energy is transferred depends on the material properties, temperature difference, cross-sectional area, and thickness of the material.
Key Concepts:
- Heat flows from the hot region to the cold region of the material.
- Rate of energy transfer is proportional to the temperature difference and cross-sectional area, and inversely proportional to the thickness.
- Thermal Conductivity (k): Measures how well a material conducts heat. Higher \(k\) → faster conduction.
Mathematical Expression (Fourier’s Law):
\( \dfrac{Q}{t} = \dfrac{k A (T_{hot} – T_{cold})}{L} \)
- \( \dfrac{Q}{t} \): Rate of heat transfer (W)
- \( k \): Thermal conductivity of the material (W/m·K)
- \( A \): Cross-sectional area perpendicular to heat flow (m²)
- \( L \): Thickness of the material (m)
- \( T_{hot} – T_{cold} \): Temperature difference across the material (K)
Example :
A metal rod of length 0.4 m and cross-sectional area \( 2 \times 10^{-4} \, m^2 \) has thermal conductivity \( k = 150 \, W/m·K \). One end is at 100 °C and the other at 25 °C. Find the rate of heat transfer by conduction.
▶️ Answer/Explanation
\( \dfrac{Q}{t} = \dfrac{k A (T_{hot} – T_{cold})}{L} \)
\( = \dfrac{150 \times 2 \times 10^{-4} \times (100 – 25)}{0.4} \)
\( = \dfrac{150 \times 2 \times 10^{-4} \times 75}{0.4} \)
\( = \dfrac{2.25}{0.4} \approx 5.625 \, W \)
Answer: Rate of heat transfer ≈ 5.63 W.
Example :
A wall of thickness 0.2 m and area 5 m² has thermal conductivity \( k = 0.8 \, W/m·K \). The inside surface is at 20 °C and the outside at 0 °C. Find the rate of heat conduction through the wall.
▶️ Answer/Explanation
\( \dfrac{Q}{t} = \dfrac{k A (T_{hot} – T_{cold})}{L} \)
\( = \dfrac{0.8 \times 5 \times (20 – 0)}{0.2} \)
\( = \dfrac{0.8 \times 5 \times 20}{0.2} = \dfrac{80}{0.2} = 400 \, W \)
Answer: Rate of heat transfer = 400 W.