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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

Sagot :

xero099

Answer:

The heat input from the combustion phase is 2000 watts.

Explanation:

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

[tex]\eta = \frac{\dot W}{\dot Q}[/tex] (1)

Where:

[tex]\dot Q[/tex] - Heat input, in watts.

[tex]\dot W[/tex] - Power output, in watts.

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

[tex]\eta = \frac{\dot W}{\dot Q}[/tex]

[tex]\dot Q = \frac{\dot W}{\eta}[/tex]

[tex]\dot Q = \frac{600\,W}{0.3}[/tex]

[tex]\dot Q = 2000\,W[/tex]

The heat input from the combustion phase is 2000 watts.

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