username970964
Badges: 14
Rep:
?
#1
Report Thread starter 3 years ago
#1

  1. Show that the change of entropy delta S in a metal block of mass m and specific heat capacity C subjected to a change in temperature from T1 to T2 is 􏰂delta S 􏰁=mCloge(T2 /T1 ).

Very stuck, I've looked up books and the internet and tried to work backwards but have no idea why there's a loge (the e is a small e). The closest idea I got was the ideal thermodynamic efficiency n = 1-T2/T1 which kind of looks like the (T2/T1) in the equation of the question. How do I even start this?
0
reply
Eimmanuel
Badges: 13
Rep:
?
#2
Report 3 years ago
#2
(Original post by Airess3)
  1. Show that the change of entropy delta S in a metal block of mass m and specific heat capacity C subjected to a change in temperature from T1 to T2 is 􏰂delta S 􏰁=mCloge(T2 /T1 ).


Very stuck, I've looked up books and the internet and tried to work backwards but have no idea why there's a loge (the e is a small e). The closest idea I got was the ideal thermodynamic efficiency n = 1-T2/T1 which kind of looks like the (T2/T1) in the equation of the question. How do I even start this?
You need to start from the "definition" of infinitesimal entropy change
dS during an infinitesimal reversible process at absolute temperature T

 dS = \dfrac {dQ}{T}

loge is natural log.

Spoiler:
Show


This problem is found in the physics textbook that you have. It is a solved example in chapter 20.

0
reply
langlitz
Badges: 17
Rep:
?
#3
Report 3 years ago
#3
(Original post by Airess3)
  1. Show that the change of entropy delta S in a metal block of mass m and specific heat capacity C subjected to a change in temperature from T1 to T2 is 􏰂delta S 􏰁=mCloge(T2 /T1 ).

Very stuck, I've looked up books and the internet and tried to work backwards but have no idea why there's a loge (the e is a small e). The closest idea I got was the ideal thermodynamic efficiency n = 1-T2/T1 which kind of looks like the (T2/T1) in the equation of the question. How do I even start this?
 dS=\frac{dQ}{T}
So the energy needed to heat a block of mass m, with specific heat capacity C by an amount dT is
dQ=mC dT
thus we have that
dS=\frac{mC}{T}dT
\int_{S_1}^{S_2} dS=\int_{T_1}^{T_2} \frac{mC}{T}dT

I'll leave you to do the integral
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Have you experienced financial difficulties as a student due to Covid-19?

Yes, I have really struggled financially (56)
17.28%
I have experienced some financial difficulties (92)
28.4%
I haven't experienced any financial difficulties and things have stayed the same (121)
37.35%
I have had better financial opportunities as a result of the pandemic (44)
13.58%
I've had another experience (let us know in the thread!) (11)
3.4%

Watched Threads

View All