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
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
1 answer
2 answers