Specific heat capacity and specific latent heat
Section: Temperature | Syllabus: Cambridge A Level Physics (9702)
Specific Heat Capacity (c)
Specific heat capacity is the quantity of thermal energy required to raise the temperature of 1 kg of a substance by 1 K (or 1 ^).
Δ Q = mcΔθUnits: J kg^-1 K^-1. Note: Δθ in K and °C are numerically equal.
Real-World Application: Thermal InertiaWater has a very high c ≈ 4200 J kg^-1 K^-1, while concrete is only ≈ 880 J kg^-1 K^-1. This explains why swimming pools stay cool while the surrounding ground gets burning hot under the same sun.
Worked Example: Lead Block HeatingQuestion: A 250 g lead block is heated from 20 ^ with 2.6 kJ of energy. Find the final temperature (c_lead = 130 J kg^-1 K^-1).
Solution1. Δθ = Δ Q / mc = 2600 / (0.25 × 130) = 80 ^.
2. Final T = 20 + 80 = 100 ^.
Experimental Skills: Determining c
Method for Aluminium1. Use a solid block with holes for a thermometer and a 12 V immersion heater.
2. Add oil to the holes to ensure good thermal contact.
3. Measures V, I, t, m, and Δθ. Calculate c = VIt / mΔθ.
4. Tip: Cooling the block below room temp before starting reduces net heat loss.
Specific Latent Heat (L)
Specific latent heat is the energy per unit mass required to change the state of a substance with no change in temperature. 'Latent' means hidden because the energy goes into changing potential energy (breaking bonds) rather than kinetic energy.
Q = mL- Fusion (L_f): Solid to liquid at melting point.
- Vaporisation (L_v): Liquid to gas at boiling point.
Graph of Temperature vs. Energy. Horizontal plateaus represent latent heat (fusion and vaporisation) where T is constant. Sloped regions represent specific heat capacity where particles' E_K is increasing.
Thermal Energy Transfer on Mixing
In a closed system, Energy Gained by Cooler Substance = Energy Lost by Hotter Substance.
Complex Example: Ice in WaterQuestion: 35 g of ice at -18 ^ is added to 100 g of water at 22 ^. Find the final equilibrium temperature.
Solution Breakdown1. Energy for ice to reach 0 ^ = mc_iceΔθ = 0.035 × 2000 × 18 = 1.26 kJ.
2. Energy to melt ice = mL_f = 0.035 × 334 = 11.69 kJ.
3. Energy for melted ice to reach θ = 0.035 × 4200 × θ = 0.147θ kJ.
4. Energy lost by water = 0.10 × 4200 × (22 - θ) = 9.24 - 0.42θ kJ.
5. Equate: 1.26 + 11.69 + 0.147θ = 9.24 - 0.42θ. Final θ = 6.6 ^.
Interactive revision notes, videos and practice questions load below.