Specific Heat Capacity — q = mcΔT Worked Examples
How much energy does it take to heat a cup of water, and why does the same flame heat metal faster than it heats water? Both questions come down to one property — specific heat capacity — and one formula. This guide gives the formula, what each symbol means, and four worked examples: heating a sample, finding an unknown material's specific heat, a calorimetry mixing problem, and a calorie-to-joule conversion.
The formula
q is the heat energy transferred (in joules, J), m is the mass of the substance (in grams, g), c is its specific heat capacity (in J g⁻¹ °C⁻¹, the energy needed to raise 1 g of the substance by 1°C), and ΔT is the temperature change, Tfinal − Tinitial (in °C or K — the two scales give the same numerical ΔT, since both use degrees of equal size). The specific heat capacity of water is commonly taken as c = 4.18 J g⁻¹ °C⁻¹ in school-level work.
Worked example 1 — heating a known sample
Question: How much heat is needed to raise the temperature of 250 g of water from 25°C to 75°C?
ΔT = 75 − 25 = 50°C
q = m × c × ΔT = 250 × 4.18 × 50
250 × 4.18 = 1045; 1045 × 50 = 52,250
q = 52,250 J = 52.25 kJ
Worked example 2 — finding an unknown specific heat capacity
Question: A 100 g metal sample absorbs 836 J of heat and its temperature rises from 20°C to 40°C. Find its specific heat capacity.
ΔT = 40 − 20 = 20°C
Rearranging q = mcΔT for c: c = q ÷ (m × ΔT)
c = 836 ÷ (100 × 20) = 836 ÷ 2000
c = 0.418 J g⁻¹ °C⁻¹
Notice this metal needs roughly a tenth of the energy water needs to raise the same mass by the same temperature — metals generally have much lower specific heat capacities than water, which is why a metal spoon heats up far faster than the tea around it.
Worked example 3 — a calorimetry mixing problem
Question: 50 g of water at 80°C is mixed with 50 g of water at 20°C in an insulated container. Find the final temperature, assuming no heat is lost to the surroundings.
Heat lost by the hot water = heat gained by the cold water:
m1c(T1 − Tf) = m2c(Tf − T2)
Since both masses and both specific heat capacities are equal (same substance, same c), they cancel from both sides:
80 − Tf = Tf − 20
100 = 2Tf
Tf = 50°C
With equal masses of the same liquid, the final temperature is always the simple average of the two starting temperatures — a useful check before trusting a longer calculation.
Worked example 4 — calories to joules
Question: 500 calories of heat are supplied to 20 g of water. Find the temperature rise, taking 1 cal = 4.184 J.
q = 500 × 4.184 = 2092 J
Rearranging for ΔT: ΔT = q ÷ (m × c)
ΔT = 2092 ÷ (20 × 4.18) = 2092 ÷ 83.6
ΔT ≈ 25.02°C
Common mistakes that cost marks
- Getting ΔT the wrong way round. ΔT = Tfinal − Tinitial, not the reverse — the sign matters when heat is released rather than absorbed.
- Mixing units of mass. If c is quoted per gram, m must be in grams, not kilograms — using kilograms without adjusting c overstates the answer by a factor of 1000.
- Confusing specific heat capacity with heat capacity. Heat capacity C = m × c is a property of one particular object; specific heat capacity c is a property of the material itself, independent of how much of it you have.
- Mixing calories and joules without converting first. Every quantity in one calculation must be in the same energy unit before q = mcΔT is applied.
- Thinking a temperature difference must be converted to Kelvin. A difference of 1°C equals a difference of 1 K exactly, so ΔT is the same number in either scale — only single-point temperatures (not differences) need the +273.15 conversion.
Where this appears in exams
| Context | Typical use |
|---|---|
| Class 11 Physics (Thermal Properties of Matter) | Calorimetry, specific heat capacity, heat exchange between substances |
| Class 11 Chemistry (Thermodynamics) | Heat as a form of energy transfer, sign conventions for q |
| JEE/NEET Physics | Calorimetry numericals, mixing problems, calibration questions |
| Everyday-science reasoning | Why water heats and cools slowly compared to metals — coastal vs inland climate, cooking |
Check your own working. Solve a calorimetry problem by hand using q = mcΔT, then verify the arithmetic on the Scientific Calculator before moving to the next question.
Open the ABC Chemistry Calculator Suite →Thermal properties and thermodynamics are core Class 11–12 chapters. ABC Chemistry covers them in its Class 11–12 batches, at the Gurugram centre and online across India: abcchemistry.in.