Posted by **ikye** on Wednesday, July 3, 2013 at 4:45am.

a 0.5kg piece of metal (c=600j/kg/k) at 300c is dumped into a large pool of water at 20c. assuming the change in temperature of water to be negligible,calculate the overall change in entropy for the system (a)85.5j/k(b) 67.4j/k(c) 122.3j/k

- phy -
**Elena**, Wednesday, July 3, 2013 at 5:25am
dS= dQ/T = m•c•dT/T

After integration

ΔS=mc ln(T₂/T₁).

The temperatures must be in Kelvins.

For piece of metal

ΔS₁=0.5•600•ln(293/573) =201.2 J/K.

For the pool of water (T=const)

dS=dQ/T = > ΔS₂ = ΔQ/T = m•c•ΔT/T₂,

ΔS₂ =0.5•600•(573-293)/293 = 286.7 J/K

The overall change in entropy for the system is the sum of two entropy changes:

ΔS=ΔS₁+ΔS₂= -201.2 +286.7 = 85.5 J/K

Answer : (a) 85.5 J/K

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