Question

If a heat engine has an efficiency of 30% and its power output is 600 W, what is the rate of heat input from the combustion phase

Answer 1

The heat input from the combustion phase is 2000 watts.

Step-by-step explanation:

The energy efficiency of the heat engine (
`\eta`), no unit, is defined by this formula:

`\eta = (\dot W)/(\dot Q)` (1)

Where:

`\dot Q` - Heat input, in watts.

`\dot W` - Power output, in watts.

If we know that
`\dot W = 600\,W` and
`\eta = 0.3`, then the heat input from the combustion phase is:

`\eta = (\dot W)/(\dot Q)`

`\dot Q = (\dot W)/(\eta)`

`\dot Q = (600\,W)/(0.3)`

`\dot Q = 2000\,W`

The heat input from the combustion phase is 2000 watts.