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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, 20:09: 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, 20:51: PM
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