Example 1
Heat 0.50kg of water by 12K. Use c=(kgK)4200J and neglect losses.
Q= mcΔT=0.50×4200×12=25200J.
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Year 11 · SACE Stage 1
Specific heat capacity
Calculate energy changes without a phase transition.
Method and worked examples
For a temperature change with no phase transition, Q= mcΔT. Here Q is energy transferred to the sample in joules, m is mass in kilograms, c is specific heat capacity in (kgK)J, and ΔT is final minus initial temperature. A positive Q warms the sample; a negative Q represents energy leaving it under this sign convention. Convert grams to kilograms before substitution.
Specific heat capacity is energy per kilogram per kelvin. Whole-object heat capacity is C= mc, in KJ. If heat loss is negligible, heater energy Pt equals mcΔT. For a graph of added energy against temperature, gradient is mc, not c alone. Real heat loss means not all electrical input becomes the sample’s temperature rise.
Example 1
Example 2
A 0.20kg metal absorbs 1800J and warms by 20K without a phase change. Find c.
c=(mΔT)Q=0.20×201800=(kgK)450J.
Watch out: The increase also depends on mass and specific heat capacity: ΔT=(Qmc), when there is no phase change.