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Pythagorean
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Thermo --> Heat Pump, confirmation.
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?
[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.
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.
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.
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