- #1
goggles31
- 34
- 0
The thermal efficiency of a reversible heat engine is solely a function of the temperature of the two reservoirs.
η = f(θH,θC) = 1 - (QC/QH)
(QC/QH) = 1 - f(θH,θC)
(QC/QH) = Ψ(θH,θC)
The simplest function that can be used is T1/T2
(QC/QH) = T1/T2
In order to define the Kelvin scale we assign a value to one of the temperatures and that is the triple point of water, 273.16K. Thus any other temperature is defined as
T = 273.16(Q/Qtp)
My question is, why was the triple point of water used in this equation and can we use any other temperature?
How does this relate to the definition of Kelvin:"The kelvin, unit of thermodynamic temperature, is the fraction 1/273.16 of the thermodynamic temperature of the triple point of water"
If we use another function, say (T1)2/(T2)3, can we stilldefine the Kelvin scale?
In practice, is this equation useful? We could just measure the temperature using a sensor instead of finding the heat transfer at both reservoirs before calculating the temperature.
η = f(θH,θC) = 1 - (QC/QH)
(QC/QH) = 1 - f(θH,θC)
(QC/QH) = Ψ(θH,θC)
The simplest function that can be used is T1/T2
(QC/QH) = T1/T2
In order to define the Kelvin scale we assign a value to one of the temperatures and that is the triple point of water, 273.16K. Thus any other temperature is defined as
T = 273.16(Q/Qtp)
My question is, why was the triple point of water used in this equation and can we use any other temperature?
How does this relate to the definition of Kelvin:"The kelvin, unit of thermodynamic temperature, is the fraction 1/273.16 of the thermodynamic temperature of the triple point of water"
If we use another function, say (T1)2/(T2)3, can we stilldefine the Kelvin scale?
In practice, is this equation useful? We could just measure the temperature using a sensor instead of finding the heat transfer at both reservoirs before calculating the temperature.