IB — physicsMedium3 min read

Thermal Concepts and Heat Transfer

Temperature; internal energy; Q = mcΔT; Q = mL; conduction, convection, radiation

Tutor AI Team
Tutor AI Team

# Thermal Concepts and Heat Transfer — IB Physics

Thermal physics explores the relationship between heat, temperature, and the microscopic behaviour of particles. These concepts are foundational for understanding energy transfer in natural and engineered systems.

1. Temperature vs Heat

Temperature: A measure of the average KE of particles. SI unit: kelvin (K). T(K)=T(°C)+273T(K) = T(°C) + 273

Heat (thermal energy): Energy transferred due to a temperature difference. SI unit: joules (J).

Internal energy: Sum of all kinetic and potential energies of particles in a system.

2. Specific Heat Capacity

Q=mcΔT\boxed{Q = mc\Delta T}

cc = specific heat capacity (J kg⁻¹ K⁻¹): energy needed to raise 1 kg by 1 K.

Material cc (J kg⁻¹ K⁻¹)
Water 4186
Aluminium 897
Copper 385
Iron 449

3. Specific Latent Heat

Q=mL\boxed{Q = mL}

LfL_f = latent heat of fusion (solid ↔ liquid) LvL_v = latent heat of vaporisation (liquid ↔ gas)

During phase change: temperature stays constant, but internal energy changes (PE of particles changes).

4. Heating Curves

A heating curve (temperature vs energy/time) shows:

  1. Solid heating (temperature rises)
  2. Melting (flat section — latent heat of fusion)
  3. Liquid heating (temperature rises)
  4. Boiling (flat section — latent heat of vaporisation)
  5. Gas heating (temperature rises)

5. Heat Transfer Mechanisms

Conduction: Energy transfer through particle vibrations and collisions (mainly in solids). Metals are good conductors (free electrons).

Convection: Energy transfer by bulk movement of fluid. Hot fluid rises (less dense), cool fluid sinks. Only in fluids.

Radiation: Energy transfer by electromagnetic waves (infrared). Only mechanism that works in a vacuum.

Worked Example: Example 1

Problem

500 g of water at 20°C is heated to 100°C. Find the energy needed. (c=4186c = 4186 J/kg/K)

Q=mcΔT=0.5×4186×80=167,440Q = mc\Delta T = 0.5 \times 4186 \times 80 = 167{,}440 J = 167 kJ

Solution

Worked Example: Example 2

Problem

200 g of ice at 0°C is melted and heated to 50°C. (Lf=334,000L_f = 334{,}000 J/kg, c=4186c = 4186 J/kg/K)

Melting: Q1=mLf=0.2×334000=66,800Q_1 = mL_f = 0.2 \times 334000 = 66{,}800 J Heating: Q2=mcΔT=0.2×4186×50=41,860Q_2 = mc\Delta T = 0.2 \times 4186 \times 50 = 41{,}860 J Total: 108,660108{,}660 J ≈ 109 kJ

Solution

Worked Example: Thermal Equilibrium

Problem

100 g of copper at 200°C is dropped into 300 g of water at 20°C. Find final temperature.

Heat lost = Heat gained: mccc(200−T)=mwcw(T−20)m_c c_c (200 - T) = m_w c_w (T - 20) 0.1(385)(200−T)=0.3(4186)(T−20)0.1(385)(200 - T) = 0.3(4186)(T - 20) 38.5(200−T)=1255.8(T−20)38.5(200 - T) = 1255.8(T - 20) 7700−38.5T=1255.8T−251167700 - 38.5T = 1255.8T - 25116 32816=1294.3T32816 = 1294.3T T=25.4°CT = 25.4°C

Solution

7. Practice Questions

    1. Define specific heat capacity and specific latent heat. (2 marks)
    1. 2 kg of water at 80°C is mixed with 3 kg of water at 20°C. Find the final temperature. (3 marks)
    1. Explain why the temperature remains constant during melting. (3 marks)

    Answers

    1. SHC: energy per unit mass per unit temperature change. SLH: energy per unit mass to change phase at constant temperature.
    1. 2(4186)(80−T)=3(4186)(T−20)2(4186)(80-T) = 3(4186)(T-20). 2(80−T)=3(T−20)2(80-T) = 3(T-20). 160−2T=3T−60160 - 2T = 3T - 60. T=44°CT = 44°C.

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Summary

  • Q=mcΔTQ = mc\Delta T (heating/cooling); Q=mLQ = mL (phase change)
  • During phase change: T constant, internal energy changes
  • Conduction (solids), convection (fluids), radiation (EM waves — vacuum OK)
Tags:
#ib#physics#thermal#heat-transfer#specific-heat#latent-heat#temperature

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