Even if the working substance in a Carnot engine is not an ideal gas, it is stil

Question

Even if the working substance in a Carnot engine is not an ideal gas, it is still the case that Thotnot Tcold lcoldl for the temperatures of the environment and the heats transferred during the isothermal legs of the cycle. In fact, this defines the thermodynamic temperature scale we all take for granted (a) Use this fact to prove that for any Carnot cycle. qhotl cold (b) When measured experimentally 1366 if the "hot" temperature corresponds to the boiling point of water at 1 atm and the "cold" temperature corresponds to the freezing point of water at 1 atm. Show that this measurement implies that absolute zero (T= 0) corresponds to

Explanation / Answer

dqrev/T =qh/Th +qC/TC

since qH/Th= -qC/TC

dqrev/T = qC/TC- qC/TC=0

2. TH/TC= qH/qC=1.366

given TH= 373K

TC= TH/1.366 = 373/1.366=273.06 ( freezing point of water)

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