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Sir, you have calculated ∆S(lake) as∆S(lake)= ∆Q(lake) / T(lake)By the same formula if we calculate ∆S(iron)Then,∆S(iron) = ∆Q(iron) / T(iron) = 4837.5/500 = 9.675 KJ/KWHY there is a difference in answers.. explain please.
Yes sir change it.. Or tell it in the comments.. Bcz it may divert some students
Bcause temp of lake is constant (=285K) but temp of iron block is changing from 500K to 285K.
thank u brthr@@2shar808
For finding Q(lake) why have you taken the mass of iron?
he's calculating the heat lost by the iron
Why we took ΔSiron =m.Cavg∫▒dt/t instead of ΔSiron =m.Cavg(Tiron-Tlake)
Check out our trending course on Thermodynamics and Power Plant Engineering- bitly.ws/SUt3
Use coupon “TH-cam 12” to get “FLAT 12%” OFF at checkout.
Sir, you have calculated ∆S(lake) as
∆S(lake)= ∆Q(lake) / T(lake)
By the same formula if we calculate ∆S(iron)
Then,
∆S(iron) = ∆Q(iron) / T(iron)
= 4837.5/500 = 9.675 KJ/K
WHY there is a difference in answers.. explain please.
Yes sir change it.. Or tell it in the comments.. Bcz it may divert some students
Bcause temp of lake is constant (=285K) but temp of iron block is changing from 500K to 285K.
thank u brthr
@@2shar808
For finding Q(lake) why have you taken the mass of iron?
he's calculating the heat lost by the iron
Why we took ΔSiron =m.Cavg∫▒dt/t instead of ΔSiron =m.Cavg(Tiron-Tlake)