AQA · GCSE · Physics
Course version: 8463 · Version from Sept 2016
The following equation applies: change in thermal energy = mass × speci f…
Full specification wording
Temperature changes in a system and specific heat capacity: The following equation applies: change in thermal energy = mass × speci f ic heat capacity × temperature change ∆E = m c ∆ θ change in thermal energy, ∆E, in joules, J mass, m, in kilograms, kg specific heat capacity, c, in joules per kilogram per degree Celsius, J/kg °C temperature change, ∆θ, in degrees Celsius, °C.
ΔE = m c Δθ links the energy transferred to a temperature change.
Checked against the AQA 8463 specification (opens in a new tab) on .
- Point 2 of 3
Study notes
From the checked teaching notes for this specification point.
The idea
ΔE = m c Δθ links the energy transferred to a temperature change. Mass must be in kilograms, so 250 g becomes 0.25 kg before you substitute. Δθ is a difference, not a final reading: water taken from 20 °C to 80 °C has Δθ = 60 °C. So 0.5 kg of water heated through 60 °C needs ΔE = 0.5 × 4200 × 60 = 126 000 J. In the required practical the energy in comes from an electric heater, ΔE = P × t or a joulemeter reading, the block is insulated, and c is found from c = ΔE ÷ (m Δθ). Because some energy always escapes to the surroundings, the measured ΔE is larger than the energy that actually reached the block, so the experimental value of c comes out too high.
What earns the marks
- Correct substitution into ΔE = m c Δθ, with mass converted to kilograms.
- Using the temperature change rather than the starting or final temperature.
- The calculated answer given in joules, or in kJ with the conversion shown.
- A correct rearrangement, such as c = ΔE ÷ (m Δθ) when the specific heat capacity is the unknown.
- Identifying energy transferred from the sample to the surroundings as the main source of error in the practical.
Common mistakes and how to put them right
Substituting a mass in grams and leaving the answer a thousand times too large.
Using the final temperature, for example 80, instead of the change of 60.
Rearranging to c = ΔE × m × Δθ instead of dividing.
Quoting the energy in J/kg °C, which is the unit of c and not of ΔE.
Forgetting that a heater supplies ΔE = P × t, and reading the power as though it were the energy.
Exam tips
- Write the equation, then a line of substituted numbers, then the answer; substitution earns a mark even if the arithmetic slips.
- Convert grams to kilograms as the very first step and cross out the original figure so you cannot use it by mistake.
- If the answer to a practical question is a value of c, say whether your value is too high or too low and name the energy transfer that caused it.
Try the lesson
The start of the Temperature changes in a system and specific heat capacity: The following equation applies: change in thermal energy = mass × speci f ic heat capacity × temperature change ∆E = m c ∆ θ change in thermal energy, ∆E, in joules, J mass, m, in kilograms, kg specific heat capacity, c, in joules per kilogram per degree Celsius, J/kg °C temperature change, ∆θ, in degrees Celsius, °C. lesson, made from checked teaching notes for this specification point.
Have a go first
A 2 kg block of aluminium is heated and its temperature rises by 5 °C. The specific heat capacity of aluminium is 900 J/kg °C. Using ∆E = m c ∆θ, what is the change in thermal energy of the block?
Listen
ΔE = m c Δθ links the energy transferred to a temperature change. Mass must be in kilograms, so 250 g becomes 0.25 kg before you substitute. is a difference, not a final reading: water taken from 20 °C to 80 °C has Δθ = 60 °C. So 0.5 kg of water heated through 60 °C needs ΔE = 0.5 × 4200 × 60 = 126 000 J. In the required practical the energy in comes from an electric heater, ΔE = P × t or a joulemeter reading, the block is insulated, and c is found from c = ΔE ÷ (m Δθ). Because some energy always escapes to the surroundings, the measured ΔE is larger than the energy that actually reached the block, so the experimental value of c comes out too high. A strong answer earns marks for these points. correct substitution into ΔE = m c Δθ, with mass converted to kilograms. using the temperature change rather than the starting or final temperature. the calculated answer given in joules, or in kJ with the conversion shown. a correct rearrangement, such as c = ΔE ÷ (m Δθ) when the specific heat capacity is the unknown. Exam tip: Write the equation, then a line of substituted numbers, then the answer; substitution earns a mark even if the arithmetic slips.
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Test your understanding of Temperature changes in a system and specific heat capacity: The following equation applies: change in thermal energy = mass × speci f ic heat capacity × temperature change ∆E = m c ∆ θ change in thermal energy, ∆E, in joules, J mass, m, in kilograms, kg specific heat capacity, c, in joules per kilogram per degree Celsius, J/kg °C temperature change, ∆θ, in degrees Celsius, °C.
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