Question

plzzz... help in solving the following problem...,

the motor in a refrigerator has a power output of 200w. If the freezing compartment is at 270 k and the outside air is at 300 k, assuming ideal efficiency, calculate the maximum amount of heat that can be extracted from the freezing compartment in 10 minutes..

Answer 1

Answer on Question #42542 – Physics – Molecular Physics | Thermodynamics

Question.

The motor in a refrigerator has a power output of 200w. If the freezing compartment is at 270 k and the outside air is at 300 k, assuming ideal efficiency, calculate the maximum amount of heat that can be extracted from the freezing compartment in 10 minutes..

Given:

$$N = 200 \, W$$ $$T_1 = 270 \, K$$ $$T_2 = 300 \, K$$ $$t = 10 \, \text{min} = 600 \, \text{s}$$

Find:

$$Q = ?$$

Solution.

Energy conversion efficiency $(\eta)$ is the ratio between the useful output of an energy conversion machine and the input. Energy conversion efficiency for refrigerator is refrigerating efficiency $\varepsilon$:

$$\varepsilon = \frac{Q}{A} \rightarrow Q = \varepsilon \cdot A$$

$Q$ is amount of heat;

$A$ is work done.

By definition:

$$\varepsilon = \frac{T_1}{T_2 - T_1}$$

And $A = N \cdot t$

So, maximum amount of heat is:

$$Q = \frac{T_1}{T_2 - T_1} N \cdot t$$

Calculate:

$$Q = \frac{270}{30} \cdot 200 \cdot 600 = 1.08 \cdot 10^6 \, J = 1.08 \, \text{MJ}$$

Answer.

$$Q = \frac {T _ {1}}{T _ {2} - T _ {1}} N \cdot t = 1.08 \cdot 10^{6} J = 1.08 M J$$

http://www.AssignmentExpert.com/