Chemical Engineering - Chemical Engineering Thermodynamics - Discussion
Discussion Forum : Chemical Engineering Thermodynamics - Section 2 (Q.No. 49)
49.
A solid metallic block weighing 5 kg has an initial temperature of 500°C. 40 kg of water initially at 25°C is contained in a perfectly insulated tank. The metallic block is brought into contact with water. Both of them come to equilibrium. Specific heat of block material is 0.4 kJ.kg-1. K-1. Ignoring the effect of expansion and contraction and also the heat capacity to tank, the total entropy change in kJ.kg-1 , K-1 is
Discussion:
6 comments Page 1 of 1.
Saikiran said:
8 years ago
Constant volume process entropy is zero.
Dilip said:
8 years ago
It is perfectly isothermal, so q=0 and hence entropy is also zero.
Talew said:
7 years ago
How this 1.26 will come? Please.
John said:
4 years ago
@Talew.
It's an adiabatic system that's why.
It's an adiabatic system that's why.
Max said:
10 months ago
@All.
Here, option B is wrong.
It doesn't say it's reversible. Hence you have to account for the entropy changes of both the block and the water and add them up.
Here, option B is wrong.
It doesn't say it's reversible. Hence you have to account for the entropy changes of both the block and the water and add them up.
Aditya said:
8 months ago
By energy balance mcp(Ti−Tf) of block=mcp(Tf−Ti) of water
So, we will get Tf=303.16k
We know,
ΔStotal=ΔSBLOCK+ΔSWATER⇒Δs=mcplnTfTI + mcplnTfTi.
⇒ on substitution w.
will get Δs = −1.869 + 3.118 = 1.26.
So, we will get Tf=303.16k
We know,
ΔStotal=ΔSBLOCK+ΔSWATER⇒Δs=mcplnTfTI + mcplnTfTi.
⇒ on substitution w.
will get Δs = −1.869 + 3.118 = 1.26.
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