Heat Engine with Finite Heat Capacity- Is my answer correct?

In summary, the maximum work that can be done by a heat engine using a large block of metal as a reservoir with initial temperature T_i and constant heat capacitance C, with the ocean as a cold reservoir with constant temperature T_0, can be calculated using the formula W = C((T_i- T_0) - T_0\ln(\frac{T_i}{T_0})). This is based on the Carnot efficiency formula E(T) = \frac{T - T_0}{T} and the fact that for one cycle, efficiency E = \frac{dW}{CdT}.
  • #1
G01
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1. A heat engine is run with a large block of metal as a reservoir with initial temperature [tex]T_i[/tex] and constant heat capacitance C. The ocean is used as a cold reservoir, with constant temperature [tex]T_0[/tex]. What is the maximum work that could be done by the engine in terms of [tex]T_i T_0 [/tex] and C
2. [tex]C = \frac{dQ}{dT}[/tex] For one cycle of the engine: Efficiency [tex] E = \frac{dW}{dQ}[/tex]Here's My attempt: (Can someone please verify if my answer is correct?)
If [tex] E = \frac{dW}{dQ}[/tex] for one cycle
Then:
[tex] E = \frac{dW}{CdT}[/tex]
[tex] dW = CE(T)dT[/tex] where E(T) = formula for Carnot Efficiency = [tex]\frac{T - T_0}{T}[/tex]

Adding up the work done in every cycle for every infinitessimal change in the metal block's temp gives:

[tex] W = C\int_{T_0}^{T_i} E(T)dT [/tex]

[tex] W = C\int_{T_0}^{T_i} \frac{T - T_0}{T}dT [/tex]

[tex] W = C\int_{T_0}^{T_i} 1 - \frac{T_0}{T} dT [/tex]

[tex] W = C((T_i- T_0) - T_0\ln(\frac{T_i}{T_0})) [/tex]


Does this formula look correct?
 
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  • #2
bumping the thread...
 
  • #3
Wow..

I can't find anything wrong. I have been in trouble with this problem. (also with two-finite heat reservoir problem). I think this is correct...
 
  • #4
hi, nope.. it aint made right..will post for, if requested for..
 
  • #5

Yes, your formula looks correct. You have correctly used the Carnot efficiency formula and integrated over the temperature range. Your final formula takes into account the finite heat capacity of the metal block and the temperature difference between the hot and cold reservoirs. Well done!
 

1. How does a heat engine with finite heat capacity work?

A heat engine with finite heat capacity operates by converting heat energy into mechanical work. It does this by utilizing a source of heat, such as burning fuel, to produce motion, which can then be harnessed for work.

2. What is the main difference between a heat engine with finite heat capacity and a heat engine with infinite heat capacity?

The main difference between these two types of heat engines is the amount of heat they can absorb and convert into work. A heat engine with finite heat capacity has a limited amount of heat it can use, while a heat engine with infinite heat capacity can use an unlimited amount of heat.

3. How does the efficiency of a heat engine with finite heat capacity compare to a heat engine with infinite heat capacity?

The efficiency of a heat engine with finite heat capacity is generally lower than that of a heat engine with infinite heat capacity. This is because the finite heat capacity limits the amount of heat that can be converted into work, resulting in a lower overall efficiency.

4. Can a heat engine with finite heat capacity be used to generate electricity?

Yes, a heat engine with finite heat capacity can be used to generate electricity. This is commonly seen in power plants, where heat from burning fuel is used to produce steam, which then turns a turbine to generate electricity.

5. Are there any disadvantages to using a heat engine with finite heat capacity?

One major disadvantage of a heat engine with finite heat capacity is its limited efficiency. Additionally, these types of engines require a constant source of heat, which can be costly and environmentally damaging. There is also the risk of overheating and damaging the engine if it is not properly managed.

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