Thermo -> Heat Pump, confirmation.

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SUMMARY

The discussion centers on the calculation of the minimal temperature a fridge can reach when using a heat pump, given a power input of 100 W and an external heat influx of 350 W. The relevant equations are established, specifically the relationship between heat transfer and temperature, expressed as \(\frac{q_2}{q_1} \leq \frac{T_2}{T_1}\). The participant concludes that the minimal temperature achievable is approximately 231.78 K, based on the correct interpretation of heat flow and work input. Clarifications regarding the values used in calculations were also addressed, ensuring accuracy in the application of the formulas.

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  • Understanding of thermodynamics principles, specifically heat transfer.
  • Familiarity with the Carnot cycle and heat pump efficiency.
  • Knowledge of basic algebra and manipulation of equations.
  • Experience with thermodynamic equations, particularly \(\frac{q_2}{q_1} \leq \frac{T_2}{T_1}\).
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  • Study the Carnot efficiency of heat pumps and refrigerators.
  • Learn about the implications of insulation on heat transfer in refrigeration systems.
  • Explore advanced thermodynamic equations and their applications in real-world scenarios.
  • Investigate the impact of varying environmental temperatures on heat pump performance.
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Students and professionals in thermodynamics, mechanical engineers, and anyone involved in refrigeration technology or heat pump systems.

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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



[tex]\frac{q_2}{q_1} \leq \frac{T_2}{T_1}[/tex]

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:

[tex]T_2 \geq \frac{q_2 T_1}{q_2} \geq \frac{350W*t(298 K)}{450W*t} \geq 231.78 K[/tex]Specific 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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