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← JEE Main 2016 · 9 Apr · Shift 1
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JEE Main 2016 · 9 Apr · Shift 1, Physics question 8

200 g water is heated from 40°C to 60°C. Ignoring the slight expansion of water, the change in its internal energy is close to (Given specific heat of water = 4184 J/kg/K) :

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Answer and worked solution

Official answer: C

ΔU≈mcΔT=0.2×4184×20=16.7\Delta U\approx mc\Delta T=0.2\times4184\times20=16.7 kJ.