Part B: "150W Engine Efficiency - Heat Given off Per Cycle

[jennypear]

An engine with an output of 150W has an efficiency of 25%. It works at 10 cycles/s.

A. How much work is done in each cycle?
work=power(delta t) = 150 W * 1s = 150 J
work =15.0 J per cycle

**got part A correct

B. How much heat is given off in each cycle?
delta internal energy = delta work + delta heat
delta intergy = 0 bc its a complete cycle
delta heat = - delta work
delta heat = 15 J

**Part B isn't coming out...please guide in the correct direction if possible.

 

[Doc Al]

Realize that what you are calling "delta heat" equals [itex]Q_{in} - Q_{out}[/itex]. What you need to find is [itex]Q_{out}[/itex]. Also recall the definition of efficiency: [itex]\eta = W_{out}/Q_{in}[/itex].

 

[thepatientmental]

In order to calculate the heat given off per cycle, we need to first determine the amount of work done in each cycle. As we established in part A, the work done in each cycle is 15 J.

Next, we need to use the efficiency equation (efficiency = work output/energy input) to find the energy input. We know that the engine has an efficiency of 25%, so we can set up the equation as:

0.25 = 15 J/Energy input

Solving for energy input, we get:

Energy input = 60 J

This means that 60 J of energy is being input into the engine in each cycle.

Now, we can use the first law of thermodynamics (delta internal energy = delta work + delta heat) to find the heat given off per cycle. Since the engine is operating at a steady state, the delta internal energy is 0. We also know that the work done in each cycle is 15 J.

Therefore, we can set up the equation as:

0 = 15 J + delta heat

Solving for delta heat, we get:

Delta heat = -15 J

This means that 15 J of heat is being given off in each cycle.

In summary, the engine with an output of 150W, operating at 10 cycles/s, has an efficiency of 25% and gives off 15 J of heat in each cycle.