Question

A heat engine with a thermal efficiency of 45 percent rejects 500 kj/kg of heat. how much heat does it receive

Answer 1

The quantity of heat that the heat engine receives is equal to 909 kJ/kg.

Given the following data:

• Thermal efficiency = 45% = 0.45.

• Heat rejected = 500 kJ/kg.

To determine the quantity of heat that the heat engine receives:

How to calculate thermal efficiency.

Mathematically, thermal efficiency is given by this formula:

$Thermal \;efficiency=\frac{Q_s-Q_r}{Q_s}$

Where:

• Qs is the heat received or supplied.

• Qr is the heat rejected.

Substituting the given parameters into the formula, we have;

$0.45=\frac{Q_s-500}{Q_s}\\\\0.45Q_s=Q_s-500\\\\Q_s-0.45Q_s=500\\\\0.55Q_s=500\\\\Q_s =\frac{500}{0.55}$

Qs = 909 kJ/kg.

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