Overview
This lecture covers key problem-solving techniques from Chapter 14, focusing on heat transfer concepts such as specific heat, calorimetry, phase changes, conduction, and radiation.
Calculating Heat Required (Q = mcΔT)
- Heat required to raise the temperature of a 20 kg iron vat from 10°C to 90°C: Q = mcΔT; m = 20 kg, c (iron) = 450 J/kg°C, ΔT = 80°C.
- Calculation: Q = 7.2 × 10⁵ J for the empty vat.
- For a filled vat: Add heat for 20 kg of water, c (water) = 4186 J/kg°C, ΔT = 80°C, Q = 6.7 × 10⁶ J.
- Total heat for vat with water: Q = 7.4 × 10⁶ J.
Calorimetry & Final Temperature
- Mixing 200 cm³ of tea (95°C) with a 150 g glass cup (25°C): Use conservation of energy, ∑Q = 0.
- Convert tea volume to mass: m = 0.20 kg (density of water = 1000 kg/m³).
- Use Q = mcΔT for both tea and cup with respective specific heats.
- Solve for final temperature; result is 86°C.
Finding Specific Heat via Calorimetry
- A hot sample placed in 0.4 kg water (10°C) in a 0.2 kg aluminum cup; final temperature = 30.5°C.
- Use ∑Q = 0 with three terms: sample, water, aluminum.
- Rearrange for unknown specific heat of the sample.
- Plugging in values, calculated specific heat: c = 497 J/kg°C.
Energy Removed by Freezer for Phase Change
- To freeze 1.5 kg water at 20°C into ice at –12°C: three Q terms (cooling water to 0°C, freezing, cooling ice to –12°C).
- Use respective specific heats and latent heat.
- Total energy removed: Q = –6.6 × 10⁵ J (negative sign shows heat leaves system).
Thermal Equilibrium with Ice and Tea
- 0.5 kg ice (–10°C) in 3 kg tea (20°C): calculate final temperature and phase.
- Ice needs three Q terms (warming, melting, warming water).
- Tea needs one Q term (cooling).
- Solve for final temperature: T = 5.1°C; final phase is liquid water.
Heat Transfer by Conduction
- Rate of heat flow: Q/t = kA(ΔT)/d, where k is thermal conductivity.
- For glass: k = 0.84 W/m°C, area = 3 m², ΔT = 1°C, thickness = 3.2 mm.
- Q/t calculated as 788 W.
Heat Loss by Radiation
- Rate of heat loss: Q/t = εσA(T₁⁴ – T₂⁴), T in Kelvin.
- Athlete (34°C, 1.5 m², ε = 0.7) in a 15°C room.
- Convert temperatures to Kelvin before calculation.
- Calculated rate of heat loss: 120 W.
Key Terms & Definitions
- Specific heat (c) — amount of heat to raise 1 kg substance by 1°C.
- Latent heat of fusion — heat required to change 1 kg from solid to liquid at constant temperature.
- Calorimetry — method to measure heat transfer using thermal equilibrium.
- Thermal conductivity (k) — material property indicating ability to conduct heat.
- Radiation — heat transfer via electromagnetic waves, calculated with Stefan-Boltzmann law.
Action Items / Next Steps
- Review tables of specific heat and conductivity values.
- Practice additional problems from Chapter 14.
- Prepare for Chapter 15: thermodynamics laws.