Consider a process in which 100 kj of energy is transferred reversibly and isothermally as heat to a large block of copper. calculate the change in entropy of the block if the process takes place at (i) 0 °c, (ii) 50 °c.

a) at 0°C the entropy change is ΔS=0.366 kJ/K

b) at 50°C the entropy change is ΔS= 0.309 kJ/K

Explanation:

from the second law of thermodynamics

ΔS ≥ ∫ dQ /T

where ΔS = change in entropy , Q= heat exchanged with the surroundings

and T= absolute temperature

- For an isothermal process ∫ dQ /T= 1/T* ∫ dQ = Q/T

- For an reversible process the ΔS = ∫ dQ /T

thus

ΔS= Q/T

replacing values

a) 0°C= 273 K

ΔS= Q/T = 100 kJ /273K = 0.366 kJ/K

ΔS=0.366 kJ/K

b) 50°C= 50 + 273K = 323K

ΔS= Q/T = 100 kJ /323K = 0.309 kJ/K

ΔS= 0.309 kJ/K

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.