A steel cable of diameter 3.0 cm supports a load of 2.0kN. What is the fractional length increase of the cable compared with the
1 answer:
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Answer:
The change in internal energy of the system is -17746.78 J
Explanation:
Given that,
Pressure
Remove heat
Radius = 0.272 m
Distance d = 0.163 m
We need to calculate the internal energy
Using thermodynamics first equation
...(I)
Where, dU = internal energy
Q = heat
W = work done
Put the value of W in equation (I)
Where, W = PdV
Put the value in the equation
Hence, The change in internal energy of the system is -17746.78 J
Answer:
a)
b)
c) 0 J/K
d)S= 61.53 J/K
Explanation:
Given that
T₁ = 745 K
T₂ = 101 K
Q= 7190 J
a)
The entropy change of reservoir 745 K
Negative sign because heat is leaving.
b)
The entropy change of reservoir 101 K
c)
The entropy change of the rod will be zero.
d)
The entropy change of the system
S= S₁ + S₂
S = 71.18 - 9.65 J/K
S= 61.53 J/K