Carnot Heat Engine, thermodynamics

[theown1]

A carnot engine has an efficiency of 59% and performs 2.5x10⁴ J of work in each cycle. a) how much heat does the engine extract from its heat source in each cycle?
b) suppose the engine exhausts heat at room temp (20^oC) what is the temperature of the heat source?



I know that the efficiency of a heat engine is e=W/Q_h or for a carnot engine e =1-T_c/T_h



3. solution
I managed to find the solution for the first part which was 0.59=W/T_h(W=2.5x10⁴J)
and I just solved for T_h which came out to be 4.2x10⁴J

but I'm not sure how I go about solving for part b, I was thinking that since it said the engine exhausts heat at room temp, so does that mean that T_h-T_c=20? or do I have to convert that temperature to internal energy and solve? I'm not sure I'm pretty confused on what it means, can anyone help

 

[RTW69]

I think what this means is Tc=20 C, efficiency stays the same what is the new Th. See if that works for you.

 

[theown1]

ya that worked, thanks!

 

[Andrew Mason]

I managed to find the solution for the first part which was 0.59=W/T_h(W=2.5x10⁴J)
and I just solved for T_h which came out to be 4.2x10⁴J

Careful. I think you meant Qh not Th.

but I'm not sure how I go about solving for part b, I was thinking that since it said the engine exhausts heat at room temp, so does that mean that T_h-T_c=20? or do I have to convert that temperature to internal energy and solve? I'm not sure I'm pretty confused on what it means, can anyone help

First convert temperatures to Kelvin. What is 20C in K?

Look at your equation for efficiency (as a function of Th and Tc). You have Tc (in Kelvin), and you have the efficiency so you can easily find Th.

AM

 

[rwisz]

?I can provide a response to your question. First, let's understand the Carnot heat engine and thermodynamics. A Carnot heat engine is a theoretical engine that operates between two heat reservoirs at different temperatures and converts heat energy into work. The efficiency of a Carnot engine is given by the formula e = 1 - Tc/Th, where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir.

Now, let's solve the given problem. We know that the efficiency of the engine is 59%, which means that 59% of the heat energy extracted from the hot reservoir is converted into work. Therefore, we can write the equation as 0.59 = W/Qh, where W is the work done and Qh is the heat extracted from the hot reservoir. We are also given that the engine performs 2.5x104 J of work in each cycle. Substituting these values in the equation, we get Qh = 4.2x104 J.

Now, for part b, we need to find the temperature of the hot reservoir. We are given that the engine exhausts heat at room temperature, which is 20°C or 293 K. We can use the equation e = 1 - Tc/Th to solve for Th. Substituting the known values, we get 0.59 = 1 - 293/Th. Solving for Th, we get Th = 488 K or 215°C.

In conclusion, the temperature of the hot reservoir for the given Carnot engine is 488 K or 215°C. I hope this helps to clarify your confusion.