Calculating Work Output from Thermal Efficiency of Car Engine

[WhyPrime89]

Homework Statement

The power output of a car engine running at 1900 is 600 kilowatts per hour.
How much work is done per cycle if the engine's thermal efficiency is 30.0 percent

Homework Equations

thermal efficiency=w(out)/Q(hot)

The Attempt at a Solution

It wants the answer in KJ/s so i converted the kilowatts to KJ/sec then divided by rev/s then multiplied by .3. What am i doing wrong? Any help would be appreciated.

 

[Carid]

Looks like you should be multiplying by 10/3.
The 600 kw/hr seems to be the power output after 70% of the initial energy has been lost.

 

[thomate1]

Your approach is correct, but your units may be causing confusion. Let's break down the equations:

Power output (watt) = work done (joule) / time (second)
Thermal efficiency (unitless) = work output (joule) / heat input (joule)

To find the work output per cycle, we need to first convert the power output from kilowatts per hour to kilojoules per second. This can be done by dividing by 3600 seconds (1 hour = 3600 seconds). So, the power output in kilojoules per second is 600 kW / 3600 s = 0.167 kJ/s.

Next, we need to find the heat input per cycle. Since we know the thermal efficiency is 30%, we can set up the equation as follows:

0.30 = work output (joule) / heat input (joule)

Rearranging this equation, we get:

Heat input (joule) = work output (joule) / 0.30

Plugging in the power output in kilojoules per second (0.167 kJ/s) as the work output, we get:

Heat input (joule) = 0.167 kJ/s / 0.30 = 0.556 kJ/s

Therefore, the work output per cycle is 0.556 kJ/s. I'm not sure where you got the "rev/s" from, but it's not necessary in this calculation. Also, be careful with units - make sure they are consistent throughout the calculation. Hope this helps!