Thermodynamics Heat Engine

[ruiwp13]

Homework Statement

A haet engine works based on the difference of the temperature at the top and bottom of the ocean. The machine is solar powered and the average solar irradiance is 240W/m^2 and the temperature at the surface of the water is 25ºC. Calculate the temperature at the bottom of the ocean so that the machine debits a power of 0.1MW to a surface area of 10^4 m^2.

Homework Equations

Q=σ*A*T^4(surface)

The Attempt at a Solution

I am stuck in this one... I don't know if it is possible for you to give me some hint. I'm not sure if I need to find the efficiency here, but I'll probably will need it to get the heat at the bottom and maybe trough there get the temperature.

 

[jim hardy]

the machine debits a power of 0.1MW to a surface area of 10^4 m^2.

is sure a roundabout way to phrase what i take to be power balance...

Do you suppose he means "delivers useful work of 10 watts/m² at Carnot efficiency?

If so, it seems to me he's given you both efficiency and T_hot .

 

[Borek]

My bet is that is about efficiency only - you are given energy input and energy output. See if you can follow from here.

Edit: Jim was slightly faster, and we definitely said the same thing.

 

[ruiwp13]

First I want to thank you both for the answer. Yes, I concluded that it was supposed to use the efficiency and I was given Th being Tc the temperature at the bottom. It's done!

 

[DeeNos]

I am happy to provide some guidance on this problem. First, it is important to understand the basic principles of thermodynamics, specifically the laws of thermodynamics and how they apply to heat engines. The first law states that energy cannot be created or destroyed, only transferred or converted from one form to another. The second law states that the total entropy of a closed system cannot decrease over time.

In the case of a heat engine, it works by converting heat energy into mechanical work. In this scenario, the heat engine is using solar energy as the heat source and the temperature difference between the top and bottom of the ocean as the driving force. To calculate the temperature at the bottom of the ocean, we can use the equation for the efficiency of a heat engine:

Efficiency = 1 - (T_cold / T_hot)

Where T_cold is the temperature at the bottom of the ocean and T_hot is the temperature at the surface of the water. We can rearrange this equation to solve for T_cold:

T_cold = T_hot / (1 - Efficiency)

Now, we are given the power output of the heat engine (0.1 MW) and the surface area (10^4 m^2). We can use the equation for power (P) to calculate the heat input (Q) to the heat engine:

P = Q/t

Where t is the time period over which the power is being output. Since we are not given a specific time period, we can assume it is over a period of 1 second. So, we can rewrite the equation as:

Q = P * t

Now, we can use the equation for heat transfer to calculate the heat input (Q) to the heat engine:

Q = σ * A * T^4

Where σ is the Stefan-Boltzmann constant, A is the surface area, and T is the temperature. Rearranging this equation to solve for T, we get:

T = (Q / (σ * A))^1/4

Now, we can plug in the values we have and solve for T_hot:

T_hot = (0.1 MW * 1 s) / (5.67 x 10^-8 W/m^2K^4 * 10^4 m^2))^1/4

T_hot = 305.9 K = 32.9 ºC

Finally, we can plug this value into our