q = m × c × ΔT — fill in every field except the one you want; the empty field is solved for.
Specific heat (J/g·°C, about 25 °C):
J/(g·°C)
Heat energy (q) = 62.76 kJ
- Convert ΔT: 60 °C =
60 K - Rearrange:
q = m × c × ΔT - Substitute:
(250 g) × (4.184 J/(g·°C)) × (60 K)=62,760 J
How it works
The calculator works in joules, grams and kelvin (a temperature change of 1 °C is the same as 1 K), solves q = m × c × ΔT for the empty field, and converts back to your units. Enter ΔT as final minus initial: a negative ΔT gives a negative q, meaning heat is released.
The presets are typical specific heat capacities near 25 °C. Water's value, 4.184 J/(g·°C), is also the definition of the thermochemical calorie. This equation only applies while nothing melts or boils; during a change of state the temperature stays constant while heat is absorbed or released.
Frequently asked questions
- What is specific heat capacity?
- It is the energy needed to raise the temperature of 1 g of a substance by 1 °C. Water's is high, 4.184 J/(g·°C), which is why oceans warm and cool slowly.
- How much energy does it take to heat water for a cup of tea?
- Warming 250 g of water from 20 °C to 80 °C needs q = 250 × 4.184 × 60 = 62,760 J, or about 62.8 kJ, ignoring heat lost to the cup and air.
- How do I find specific heat in a calorimetry experiment?
- Measure the heat absorbed by a known mass of water (q = m c ΔT for the water), assume the sample lost the same amount of heat, then solve c = q ÷ (m ΔT) for the sample.
- Why is q negative?
- A negative q means the substance lost heat to its surroundings — its temperature fell, so ΔT is negative.
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