# Rate of Heat Output to Low-Temp Res per Cycle: 450J

• fish
In summary, the heat output to the low-temperature reservoir per cycle is 450J. The book's answer of 800J is incorrect, as the heat output should be equal to the heat input minus the heat output, which is 600J. It is possible that the book mixed up the input and output heats.
fish
A heat engine has an efficiency of 25.0% and its heat input is 600J per cycle from the high-temperature reservoir. What is the rate of heat output to the low-temperature reservoir per cycle?

E=(Qh-Qc)/Qh

.25=(600J-Qc)/600J
Qc=450J

book has the answer as 800J.
How are they getting 800J and not 450J?

Originally posted by fish
book has the answer as 800J.
How are they getting 800J and not 450J?
The book's answer makes no sense. Your reasoning is correct. Are you sure you're not mixing up input and output heats? That would explain the book's answer.

yes, your're right. It looks like I mixed up input and output heats.
Qc=600J (heat flow into cold res.)

find heat output from high-temp res. which would be
Qh, so solving for Qh you get 800J

## 1. What is the significance of the Rate of Heat Output to Low-Temp Res per Cycle?

The Rate of Heat Output to Low-Temp Res per Cycle is a measure of the amount of heat that is released from a system to a low-temperature reservoir in one cycle. It is an important parameter in thermodynamics, as it helps to determine the efficiency of a system.

## 2. How is the Rate of Heat Output to Low-Temp Res per Cycle calculated?

The Rate of Heat Output to Low-Temp Res per Cycle is calculated by dividing the amount of heat released from the system by the number of cycles. It is typically measured in joules (J) per cycle.

## 3. What factors can affect the Rate of Heat Output to Low-Temp Res per Cycle?

The Rate of Heat Output to Low-Temp Res per Cycle can be affected by various factors, such as the temperature difference between the system and the low-temperature reservoir, the type of material used in the system, and any external forces or constraints on the system.

## 4. How can the Rate of Heat Output to Low-Temp Res per Cycle be optimized?

To optimize the Rate of Heat Output to Low-Temp Res per Cycle, one can try to minimize any energy losses in the system, increase the temperature difference between the system and the low-temperature reservoir, and use materials with high thermal conductivity.

## 5. What is a typical value for the Rate of Heat Output to Low-Temp Res per Cycle?

The typical value for the Rate of Heat Output to Low-Temp Res per Cycle can vary greatly depending on the specific system and conditions. However, it is often in the range of a few hundred to a few thousand joules per cycle.

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