Elementary Thermodynamics Question

AI Thread Summary
The discussion revolves around a thermodynamics problem involving a freezer and the heat exchange with the surrounding air. The user successfully calculated the heat removed from the food, the energy delivered to the freezer motor, and the total heat ejected into the room. The main issue lies in determining the temperature change in the room, where the user calculated approximately 6.2 degrees Celsius, while the instructor indicated the answer should be 7.3 degrees Celsius. Despite using the formula Q = cm(delta)t correctly, there is confusion regarding the discrepancy in results. The conversation highlights the importance of significant figures and potential calculation errors in thermodynamic equations.
Steven925
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The prompt for the question I am seeking help on is:

"A freezer contains 20 kg of food with a specifit heat of 2x10^3 J/kg*c. The temperature inside the freezer is initially -5*C. The freezer motor them operates for 10 minutes, reducing the temperature to -8*C."

I have found solutions to the first three portions of the question:
a)How much heat is removed from the food during this time? The freezer operates at 400W

Q = -120,000J using Q = CM(t(f) - t(i))

b)How much energy is delivered to the freezer motor suring the 10 minute period?

P = W/T = 240,000J

c)During this time, how much total heat is ejected into the room in which the freezer is located?

Q = Q + W = 360,000J

----------------

D is what i need help on:

"Determine the temperature change in the room if the specific heat of air is 700J/Kg*C. assume there are 80 kilograms of air in the room, the volume of the air is constant, and there is no heat loss from the room"

I have tried using the standard Q = cm(delta)t:

360,000J = 700J/Kg*c * 80 (delta)t

but that comes out to approximately 6.2 degrees celcius. My instructor says the answer is 7.3*C and I'm unsure of what I may be doing incorrectly. Any guidance would be greatly appreceiated.

Thanks!
 
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hrmmm. I tried doing mcAt = mcAt...

but it doesn't seem to work. lol
 
Steven925 said:
I have tried using the standard Q = cm(delta)t:

360,000J = 700J/Kg*c * 80 (delta)t

but that comes out to approximately 6.2 degrees celcius. My instructor says the answer is 7.3*C and I'm unsure of what I may be doing incorrectly. Any guidance would be greatly appreceiated.

Thanks!
I don't see anything wrong with your method. I get 6.4 deg. C which to the correct significant figures should be 6 deg. C.

AM
 
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