Question

Are enthalpy and entropy affected by temperature?

Answer 1

Yes. In fact, they are state functions of temperature.

`color(blue)(DeltaH(T) = int_(T_1)^(T_2) C_P dT) = C_P(T_2 - T_1)`

`color(blue)(DeltaS(T) = int_(T_1)^(T_2) C_P/T dT) = C_Pln|T_2/T_1|`

If we assume that the constant-pressure heat capacity `C_P` is constant (which is a pretty decent assumption for small temperature ranges), then we can pull it out of the integral and focus on integrating the temperature.

(Normally though, `C_P = C_P(T)`.)

From this we can say that enthalpy and entropy change proportionally to a change in temperature.

This is derived below.

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The change in enthalpy with respect to temperature at a constant pressure is the definition of the constant-pressure heat capacity.

`\mathbf(((delH)/(delT))_P = (delq_p)/(delT) = C_P)`

where `q_p` is heat flow at a constant pressure.

From this, we can write at a constant pressure:

`dH = delq_p = C_PdT`

Now integrate everything to get:

`color(green)(DeltaH = q_p = int_(T_1)^(T_2) C_PdT) = C_P(T_2 - T_1)`

Because of that relationship between `q_p` and `DeltaH`, notice how we can relate two similar equations, provided the heat flow is reversible:

`DeltaH = q_(p,rev)`

`color(green)(DeltaS = q_(p,rev)/T)`

Using the above green relationships, we get:

`color(blue)(DeltaS = int_(T_1)^(T_2) C_P/T dT)`

Finally, since both entropy and enthalpy are functions of temperature, we emphasize it by writing:

`DeltaH = DeltaH(T)`

`DeltaS = DeltaS(T)`