Engineering
Physics
Calorimetry New
Question
100 g of water is heated from 30°C to 50°C. Ignoring the slight expansion of the water, the change in its internal energy is (specific heat of water is 4184 J/kg/K) :
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Solution
ΔQ = M,S, ΔT
= 100 × 10–3 × 4.184 × 20 = 8.4 × 103
ΔQ = 84 kJ, ΔW = 0
ΔQ = ΔV + ΔW
∴ ΔV = 8.4 kJ.