   Chapter 12, Problem 63AP

Chapter
Section
Textbook Problem

A 1 500-kW heat engine operates at 25% efficiency. The heat energy expelled at the low temperature is absorbed by a stream of water that enters the cooling coils at 20.°C. If 60. L flows across the coils per second, determine the increase in temperature of the water.

To determine
The increase in temperature of the water.

Explanation

Section1:

To determine: The heat energy expelled at the lower temperature.

Answer: The heat energy expelled at the lower temperature is 4.5×103kJ .

Explanation:

Given Info:

The work output of the given engine per second is 1500kJ .

The efficiency of the engine is 25% .

Formula to calculate the energy expelled is,

Qc=Weng(1e1)

• Weng is the work output of the engine
• e is the efficiency of the engine

Substitute 1500kJ for Weng and 0.25 for e to find the energy expelled,

|Qc|=(1500kJ)(10.251)=4.5×103kJ

Thus, the heat energy expelled at the lower temperature is 4.5×103kJ .

Conclusion:

From section 1,

The heat energy expelled at the lower temperature is 4.5×103kJ

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