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## Main Question or Discussion Point

The thermal efficiency of a reversible heat engine is solely a function of the temperature of the two reservoirs.

η = f(θ

(Q

(Q

The simplest function that can be used is T

(Q

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/Q

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:"

If we use another function, say (T

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(θ

_{H},θ_{C}) = 1 - (Q_{C}/Q_{H})(Q

_{C}/Q_{H}) = 1 - f(θ_{H},θ_{C})(Q

_{C}/Q_{H}) = Ψ(θ_{H},θ_{C})The simplest function that can be used is T

_{1}/T_{2}(Q

_{C}/Q_{H}) = T_{1}/T_{2}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/Q

_{tp})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 (T

_{1})^{2}/(T_{2})^{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.