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1 eV = 1 × 1.60218 × 10-19 ÷ 1 = 1.60218 × 10-19 J.
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
Quoted from Specific Heat Capacity — q = mcΔT Worked Examples.
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.
Quoted from Specific Heat Capacity — q = mcΔT Worked Examples.
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