Chemistry and laboratory
Specific heat calculator (q = mcΔT)
Solve q = mcΔT for heat, mass, specific heat capacity or final temperature, with common materials and unit conversion.
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Result
- Heat absorbed q
- 12.6 kJ
- Temperature change ΔT
- 60.0 K (= °C)
- Heat in kilocalories
- 3.00 kcal
Working
- q = m c ΔT, ΔT = T₂ − T₁
- q = m c ΔT = 50 g × 4.184 J/(g·K) × 60 K = 12552 J
- No phase change (melting, boiling) may happen in the range, and c is taken as constant. A positive q means heat flows into the material.
- Calories are thermochemical calories (1 cal = 4.184 J exactly).
- Rounded to 3 significant figures.
Method and limits
Choose the unknown and fill in the rest. A positive q means heat flows into the material (it warms); a negative q means it gives heat out. A change of 1 °C equals a change of 1 K.
q = m c ΔT ΔT = T(final) − T(initial) T(final) = T(initial) + q ÷ (m c)
Specific heat capacities for the preset materials are at about 25 °C, from molar heat capacities in the CRC Handbook and the NIST Chemistry WebBook divided by the CIAAW molar masses (for example aluminium: 24.20 J/(mol·K) ÷ 26.982 g/mol = 0.897 J/(g·K)). Textbook problems often use 4.184 J/(g·K) for water, the value that defines the thermochemical calorie.
Limits
There must be no change of phase (melting, boiling) between the two temperatures; that needs the latent heat as well. c is taken as constant, which is good to about 1 % for water between 0 and 100 °C. Heat lost to the surroundings or the container is not included. Answers keep the significant figures of the least precise number you enter (at least 3, at most 6).
Worked example
Warming 50.0 g of water from 20.0 °C to 80.0 °C with c = 4.184 J/(g·K): q = 50.0 g × 4.184 J/(g·K) × 60.0 K = 12 552 J = 12.6 kJ (3.00 kcal).
References
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