Calculating Change in Entropy: 1.124J Heat Added to Gas at 300K [Solved]

[boburd21]

[SOLVED] Quick Entropy Problem

Homework Statement

1.124J of heat is added to a gas at 300K causing the temperature to increase to 600K. What is the change in entropy of the gas?

Homework Equations

(Delta)S=Q/T

The Attempt at a Solution

I know the simple equation i need. all i need to know is whether to use the Tfinal or the Tinitial in the equation.
Thanks

 

[G01]

You are going to need to use the more general version of the equation you posted:

$$\Delta S = \int \frac{dQ}{T}$$

Start the problem like so: Get dQ in terms of dT and then integrate with respect to T. What will your integral's bounds be in this case?

 

[boburd21]

Im in a non-calculus based physics class. Any ideas what the teacher would want in this case if integrating is not an option?

 

[G01]

O ok, well, I would assume you would use the initial temperature of the system, but you may want to check with your teacher, since I am not completely sure.

 

[boburd21]

thanks

 

[Andrew Mason]

Im in a non-calculus based physics class. Any ideas what the teacher would want in this case if integrating is not an option?

If you use either the initial or final temperature, you will be wrong. If you are in a non-calculus class, your teacher should either be giving you the general solution to the integral (which I expect is the case) or not be giving you these kinds of questions. The correct solution to the integal is:

$$\Delta S = \int_{T_i}^{T_f}\frac{dQ}{T} = \int_{T_i}^{T_f}\frac{mC_vdT}{T} = mC_v(\ln{T_f} - \ln{T_i}) = mC_v\ln{\frac{{T_f}}{{T_i}}} = \frac{\Delta Q}{\Delta T}\ln{\frac{{T_f}}{{T_i}}} = \frac{124}{300}\ln{2}$$

AM