# Are enthalpy and entropy affected by temperature?

##### 1 Answer

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**

(Normally though,

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

This is derived below.

*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

#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)#