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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 (Take,specific heat of water is 4184 J/Kg/K).     Personally I want to know what is slight expansion and anotherly is the work done here is zero ,if yes how??
Asked by vishakhachandan026 | 02 Apr, 2019, 08:09: PM
expansion coefficient of water γ = 2×10-4 / °C

hence change in volume = V0 γ ΔT = 100× 2×10-4 × 20 = 0.4 cc

hence change in volume is from 100 cc to 100.4 cc

First law of thermodynamics :  ΔQ = ΔU + ΔW  .....................(1)

where ΔQ = heat energy given to the system ( here system is 100 g water )
ΔU = change in internal energy,
ΔW = workdone  = pΔV  , where p is pressure and ΔV is change in volume

since ΔV≈0, workdone is zero.

hence ΔQ = ΔU = m×Cp ×ΔT  = 100×10-3 ×4184×20 = 8368 J
Answered by Thiyagarajan K | 02 Apr, 2019, 08:51: PM

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