# Calculating specific heat capacity from entropy

• approx12
In summary, the conversation discusses a problem involving calculating the specific heat capacity at constant pressure and volume using given equations and variables. It is suggested to use an equation of state to simplify the calculations.
approx12
Homework Statement
Given the entropy S of system, calculate the specific heat capacity C_V and C_p
Relevant Equations
$$C_P=T \left(\frac{\partial S}{\partial T}\right)_{N,P}$$ and
$$C_V=T \left(\frac{\partial S}{\partial T}\right)_{N,V}$$
Hey guys! I'm currently struggling with a specific thermodynamics problem.
I'm given the entropy of a system (where ##A## is a constant with fitting physical units): $$S(U,V,N)=A(UVN)^{1/3}$$I'm asked to calculate the specific heat capacity at constant pressure ##C_p## and at constant volume ##C_V##.
I know that the two are given by the following equation:
$$C_P=T \left(\frac{\partial S}{\partial T}\right)_{N,P}$$$$C_V=T \left(\frac{\partial S}{\partial T}\right)_{N,V}$$I've tried to eleminate ##U## from the equation by calculating: $$\left(\frac{\partial S}{\partial V}\right)=\frac{P}{T}=\frac{1}{3}(NU)^{1/3}V^{-2/3}$$ Solving for ##U## and plugging it back into the original equation gives me: $$S(P,V,T)=\frac{PV}{T}$$
I don't know if my steps were correct so far but what I'm now struggling with is calculating ##\left(\frac{\partial S}{\partial T}\right)_{P}## and ##\left(\frac{\partial S}{\partial T}\right)_{V}##. For me they would be both equal to $$\left(\frac{\partial S}{\partial T}\right)=-\frac{PV}{T^2}$$ But I don't think that is correct because the relationship $$C_P-C_V=\frac{TV\alpha_P^2}{\kappa_T}$$ needs to bet true.

It would be awesome if anyone could help me out with this one and point me in the right direction. Thank you!

Looks like you dropped a factor of 3 in getting to the equation ##S = \frac{PV}{T}##.

I don't know if my steps were correct so far but what I'm now struggling with is calculating ##\left(\frac{\partial S}{\partial T}\right)_{P}## and ##\left(\frac{\partial S}{\partial T}\right)_{V}##. For me they would be both equal to $$\left(\frac{\partial S}{\partial T}\right)=-\frac{PV}{T^2}$$

When calculating ##\left(\frac{\partial S}{\partial T}\right)_{P}## you can't treat ##V## as a constant. Likewise when calculating ##\left(\frac{\partial S}{\partial T}\right)_{V}## you can't treat ##P## as a constant.

One way to proceed is to find the equation of state that relates the 3 variables ##P##, ##V##, and ##T##. You can then use this to write ##S## as a function of just ##T## and ##P## which then makes it straightforward to evaluate ##\left(\frac{\partial S}{\partial T}\right)_{P}##. Or, you can write ##S## as a function of just ##T## and ##V## so that you can evaluate ##\left(\frac{\partial S}{\partial T}\right)_{V}##. I don't know if this is the best way, but it's one way.

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## What is specific heat capacity?

Specific heat capacity is the amount of heat energy required to raise the temperature of one unit of mass of a substance by one degree Celsius or Kelvin.

## What is entropy?

Entropy is a measure of the disorder or randomness of a system. It is a thermodynamic property that describes the distribution of energy in a system.

## How do you calculate specific heat capacity from entropy?

The equation for calculating specific heat capacity from entropy is:
c = TΔS/m
Where c is the specific heat capacity, T is the temperature, ΔS is the change in entropy, and m is the mass of the substance.

## What units are used for specific heat capacity and entropy?

The units for specific heat capacity are typically Joules per kilogram per degree Celsius (J/kg°C) or Joules per gram per degree Celsius (J/g°C). The units for entropy are typically Joules per Kelvin (J/K).

## Can specific heat capacity and entropy be measured experimentally?

Yes, both specific heat capacity and entropy can be measured experimentally using various methods such as calorimetry and thermodynamics. These measurements can then be used to calculate the specific heat capacity from entropy.

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