Thermo -> Heat Pump, confirmation.

AI Thread Summary
The discussion centers on calculating the minimal temperature a fridge can reach when using a heat pump, given a power input of 100 W and heat influx of 350 W due to poor insulation. The key equation used is the ratio of heat transfer, which relates the heat being pumped out (q2) and the heat entering the fridge (q1). The user attempts to clarify their assumption that q2 is equivalent to 350 W multiplied by time, leading to the conclusion that the fridge's temperature can be calculated as approximately 231.78 K. There is some confusion regarding the values used in the equations, but the final calculation is confirmed as correct. The discussion emphasizes the importance of accurately transcribing equations and understanding the relationships between heat transfer and work input.
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Thermo --> Heat Pump, confirmation.

Homework Statement



A cooling unit for a fridge has a 100 W power input. Environment temperature is 298 K. The heat which enters the fridge due to defective isolation is 350 W. What is the minimal temperature the fridge can reach?

Homework Equations



\frac{q_2}{q_1} \leq \frac{T_2}{T_1}

where q2 is the heat being pumped out of the fridge, q1 the heat being pumped into the environment, T1 the environment temperature, and T2 the fridge temperature.

q1 = q2 + w

where w is the work input on the cooling device.

The Attempt at a Solution



My key assumption that allows me to do this problem I feel very weak about. That is that q2 is equivalent to 350W*t where t is some arbitrary time. My interpretation of the problem is that that's the heat we want to remove.

With this in mind:

q1 = q2 + w = 350 W*t + 100 W*t = 450 W*t

so:

T_2 \geq \frac{q_2 T_1}{q_2} \geq \frac{350W*t(298 K)}{450W*t} \geq 231.78 KSpecific Question:

Is my fundamental assumption for q2 correct? I'm just looking for hints here, not solutions.
 
Last edited:
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q1 = q2 + w = 250 W*t + 150 W*t = 350 W*t ?
Where did you get 250 and 150 ? And in the formula the denominater should be q2 + w. But the answer is OK
 
rl.bhat said:
q1 = q2 + w = 250 W*t + 150 W*t = 350 W*t ?
Where did you get 250 and 150 ? And in the formula the denominater should be q2 + w. But the answer is OK

oops! made an error in putting it from paper to computer. Made the fix, should have been

350 W*t + 100 W*t = 450 W*t

but I transcribed the final calculation correctly (q2 + w = 450 is in the denominator)
 
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