Specific Heat Capacity, Specific Latent Heat & Thermal Insulation
When a substance is heated, the thermal energy transferred either increases the average kinetic energy of its particles (raising its temperature according to Specific Heat Capacity, ΔE = mcΔθ) or increases the potential energy of its particles by overcoming intermolecular forces during a change of state (at constant temperature according to Specific Latent Heat, E = mL). Materials with a high specific heat capacity, such as water (4,200 J/kg°C), store large amounts of thermal energy per kilogram for every degree of temperature rise. Examiners frequently test multi-step calculations combining electrical power (E = Pt) with ΔE = mcΔθ, heating curve plateaus, and the metal block Specific Heat Capacity required practical.
- Specific heat capacity formula (ΔE = mcΔθ)
- The specific heat capacity (c) of a substance is the amount of energy required to raise the temperature of 1 kg of the substance by 1°C. Formula: Change in thermal energy, ΔE (Joules, J) = Mass, m (kg) x Specific heat capacity, c (J/kg°C) x Temperature change, Δθ (°C).
- Required Practical: Determining Specific Heat Capacity
- Place a 1 kg metal block (with two holes) on a heatproof mat and wrap in thermal insulation. Insert an electric immersion heater into one hole and a thermometer (with a drop of water/oil for good thermal contact) into the other. Measure electrical energy supplied using a joulemeter, or connect an ammeter in series and voltmeter in parallel with a stopclock and use E = V x I x t. Plot temperature (y-axis) against energy/work done (x-axis); the gradient of the straight-line section is Δθ/E, so c = 1 / (m x gradient).
- Specific latent heat formula (E = mL)
- The specific latent heat (L) of a substance is the amount of energy required to change the state of 1 kg of the substance with no change in temperature. Formula: Energy for a change of state, E (Joules, J) = Mass, m (kg) x Specific latent heat, L (J/kg).
Specific Heat Capacity, Specific Latent Heat & Thermal Insulation: Wave Equation (v = fλ) & Force Vector (F = ma) Simulator
Adjust physical variables on the sliders below to see real-time changes in wave geometry, resultant forces, and acceleration.
F_net = 80 N − 25 N = 55 NAcceleration: a = F_net ÷ m = 55 ÷ 10 = 5.50 m/s²Key Concepts & Exam Definitions (6)
- Specific heat capacity formula (ΔE = mcΔθ)
- The specific heat capacity (c) of a substance is the amount of energy required to raise the temperature of 1 kg of the substance by 1°C. Formula: Change in thermal energy, ΔE (Joules, J) = Mass, m (kg) x Specific heat capacity, c (J/kg°C) x Temperature change, Δθ (°C).
- Required Practical: Determining Specific Heat Capacity
- Place a 1 kg metal block (with two holes) on a heatproof mat and wrap in thermal insulation. Insert an electric immersion heater into one hole and a thermometer (with a drop of water/oil for good thermal contact) into the other. Measure electrical energy supplied using a joulemeter, or connect an ammeter in series and voltmeter in parallel with a stopclock and use E = V x I x t. Plot temperature (y-axis) against energy/work done (x-axis); the gradient of the straight-line section is Δθ/E, so c = 1 / (m x gradient).
- Specific latent heat formula (E = mL)
- The specific latent heat (L) of a substance is the amount of energy required to change the state of 1 kg of the substance with no change in temperature. Formula: Energy for a change of state, E (Joules, J) = Mass, m (kg) x Specific latent heat, L (J/kg).
- Heating & Cooling Curves (Kinetic vs Potential Energy)
- Internal energy is the total kinetic energy and potential energy of all the particles in a system. On sloped sections of a heating graph, temperature rises because particles gain kinetic energy and move faster (ΔE = mcΔθ). On horizontal flat plateaus (melting point / boiling point), temperature is constant because energy increases the potential energy store of the particles by weakening or breaking intermolecular bonds (E = mL).
- Thermal Conductivity & Building Insulation
- Thermal conductivity is a measure of how quickly energy is transferred through a material by conduction. The rate of cooling of a building is reduced by having thick walls made from materials with a low thermal conductivity, augmented by cavity wall insulation (foam traps air pockets to stop convection), loft insulation (fibreglass), and double glazing (vacuum/air gap stops conduction and convection).
- Required Practical: Thermal Insulation (Triple Science)
- Investigate the effectiveness of different insulating materials (e.g., newspaper, bubble wrap, wool, foil) or different thicknesses of the same insulator wrapped around a beaker of boiling water with a lid. Measure temperature every 2 minutes for 15 minutes using a thermometer. Control variables: initial volume of water, starting temperature of water, thickness of different materials, and room temperature.
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
Define Specific Heat Capacity and state its standard unit.
Why is water used as the coolant in car engines and in central heating radiators?
In the SHC required practical, why do you put a few drops of oil or water into the thermometer hole in the metal block?
Define Specific Latent Heat of Vaporisation.
Explain what happens to the internal energy, kinetic energy, and potential energy of ice particles as ice melts at 0°C.
How does cavity wall insulation reduce unwanted energy transfers from a warm house?
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
An electric kettle with a power rating of 2.1 kW is filled with 1.5 kg of water at 20°C. Given that the specific heat capacity of water is 4,200 J/kg°C, calculate the minimum time taken for the water to reach its boiling point of 100°C, and state why the actual time taken is slightly longer. [5 marks]
Click “Check Mark Scheme” to compare your answer, 1. Calculate temperature change: Δθ = 100°C - 20°C = 80°C [1 mark]. 2. Calculate thermal energy required using ΔE = m x c x Δθ = 1.5 x 4,200 x 80 = 504,000 J [1 mark]. 3. Convert power from kW to W: P = 2.1 kW x 1,000 = 2,100 W [1 mark]. 4. Use E = P x t -> t = E / P = 504,000 / 2,100 = 240 seconds (or 4 minutes) [1 mark]. 5. Actual time is longer because the kettle is not 100% efficient: some thermal energy is dissipated to heat the plastic/metal kettle walls and the surrounding air [1 mark].
- Q2
A student determines the specific heat capacity of an aluminium block of mass 0.80 kg. They connect an immersion heater to a 12 V power supply; the ammeter reads 3.5 A. After 10 minutes of heating, the temperature of the block has risen from 22°C to 50°C. Calculate the specific heat capacity of aluminium from these results. [5 marks]
Click “Check Mark Scheme” to compare your answer, 1. Convert time to seconds: t = 10 x 60 = 600 s [1 mark]. 2. Calculate total electrical energy supplied: E = V x I x t = 12 x 3.5 x 600 = 25,200 J [1 mark]. 3. Calculate temperature rise: Δθ = 50°C - 22°C = 28°C [1 mark]. 4. Rearrange ΔE = m x c x Δθ for c: c = ΔE / (m x Δθ) = 25,200 / (0.80 x 28) [1 mark]. 5. c = 25,200 / 22.4 = 1,125 J/kg°C [1 mark].
- Q3
A 0.25 kg block of ice is initially at -10°C. It is heated until it completely melts into liquid water at 0°C. Given that the specific heat capacity of ice is 2,100 J/kg°C and the specific latent heat of fusion of ice is 334,000 J/kg, calculate the total thermal energy required. [4 marks]
Click “Check Mark Scheme” to compare your answer, 1. Energy to warm solid ice from -10°C to 0°C (Δθ = 10°C): ΔE₁ = m x c x Δθ = 0.25 x 2,100 x 10 = 5,250 J [1 mark]. 2. Energy to melt the 0.25 kg of ice at 0°C into water at 0°C: E₂ = m x L_f = 0.25 x 334,000 [1 mark]. 3. E₂ = 83,500 J [1 mark]. 4. Total thermal energy = 5,250 J + 83,500 J = 88,750 J (or 88.75 kJ) [1 mark].
- Q4
Explain in terms of particles and internal energy why a scald from 100°C steam is much more severe than a scald from the same mass of 100°C boiling water. [3 marks]
Click “Check Mark Scheme” to compare your answer, 1. Steam at 100°C has a much higher internal energy than the same mass of liquid water at 100°C because its particles have a much greater store of potential energy [1 mark]. 2. When 100°C steam hits the skin, it first condenses into 100°C liquid water, releasing its large specific latent heat of vaporisation (E = mL = 2,260,000 J/kg) directly into the skin before its temperature even begins to drop below 100°C [1 mark]. 3. Only after condensing does it transfer the same thermal energy (ΔE = mcΔθ) as 100°C liquid water cooling down to skin temperature [1 mark].
Common Exam Questions (4)
What is the difference between Specific Heat Capacity (c) and Specific Latent Heat (L)?
Specific Heat Capacity (ΔE = mcΔθ) applies when a substance is changing temperature without changing state; it has units of J/kg°C. Specific Latent Heat (E = mL) applies when a substance is changing state (melting/boiling) at a constant temperature; it has units of J/kg.
Why does the temperature of a substance stay completely flat (constant) while it is melting or boiling on a heating graph?
During a change of state (melting or boiling), all the thermal energy being supplied is used to break/overcome the intermolecular forces of attraction between the particles, increasing their internal potential energy rather than their average kinetic energy. Because temperature is a measure of the average kinetic energy of particles, the temperature remains constant until the state change is complete.
What is the difference between specific latent heat of fusion and specific latent heat of vaporisation?
Specific latent heat of fusion (L_f) is the energy required to change 1 kg of a substance from a solid to a liquid at its melting point (without a temperature change). Specific latent heat of vaporisation (L_v) is the energy required to change 1 kg of a substance from a liquid to a gas (vapour) at its boiling point.
Why is the experimentally measured value of Specific Heat Capacity in the lab usually higher than the true databook value?
In the laboratory, some of the thermal energy from the immersion heater is dissipated (lost) to the surrounding air and bench rather than heating the metal block. Therefore, more electrical energy (E) has to be supplied to achieve a given temperature rise (Δθ); since c = E / (mΔθ), a larger numerator E gives a calculated value of c that is higher than the true value.
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