Answer:
ΔS total ≥ 0 (ΔS total = 0 if the process is carried out reversibly in the surroundings)
Explanation:
Assuming that the entropy change in the aluminium bar is due to heat exchange with the surroundings ( the lake) , then the entropy change of the aluminium bar is, according to the second law of thermodynamics, :
ΔS al ≥ ∫dQ/T
if the heat transfer is carried out reversibly
ΔS al =∫dQ/T
in the surroundings
ΔS surr ≥ -∫dQ/T = -ΔS al → ΔS surr ≥ -ΔS al = - (-1238 J/K) = 1238 J/K
the total entropy change will be
ΔS total = ΔS al + ΔS surr
ΔS total ≥ ΔS al + (-ΔS al) =
ΔS total ≥ 0
the total entropy change will be ΔS total = 0 if the process is carried out reversibly in the surroundings
120n
since the speed is doubled, her force is doubled
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Answer:
Work out = 28.27 kJ/kg
Explanation:
For R-134a, from the saturated tables at 800 kPa, we get
= 171.82 kJ/kg
Therefore, at saturation pressure 140 kPa, saturation temperature is
= -18.77°C = 254.23 K
At saturation pressure 800 kPa, the saturation temperature is
= 31.31°C = 304.31 K
Now heat rejected will be same as enthalpy during vaporization since heat is rejected from saturated vapour state to saturated liquid state.
Thus, 
= = 171.82 kJ/kg
We know COP of heat pump
COP = 
= 
= 6.076
Therefore, Work out put, W = 
= 171.82 / 6.076
= 28.27 kJ/kg
