How Fast Does Water Cool in a Polystyrene Box?

AI Thread Summary
The discussion focuses on calculating the rate of temperature drop of water in a cubic polystyrene box. The initial calculations yield a power loss of 0.00429W and an energy requirement of 273J, resulting in an estimated cooling time of approximately 636,364 seconds. Participants express uncertainty about the accuracy of the answers, particularly regarding the surface area and units of thermal conductivity. Clarifications confirm that the units for thermal conductivity are indeed correct. The conversation highlights the complexities involved in thermal calculations and the importance of unit consistency.
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Homework Statement


A cubic polystyrene box contains a litre of water at 90°C. It is 10cm wide and has a wall thickness of of 5mm. Room temperature is 25°C and the conductivity of polystyrene is 0.033W/m°C.
What is the rate of temperature drop of the water?

Homework Equations



P= (KAΔT)/diameter
E=mCΔT

The Attempt at a Solution



P= ((0.033)(0.000001)(65))/(0.005)=0.00429W
E=(1)(4.2)(65)=273
t=E/P, therefore:
t=636363.6364

ΔT/t=1.0214*10^(-6) °C/s




Not really sure if i did this right, and i feel really uncomfortable with my answers because
they're so small. especially the time. I feel i did something wrong with the Area.
 
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What is the surface area of the box?
 
Your units of thermal conductivity are incorrect. It should be per m2.
 
Chestermiller said:
Your units of thermal conductivity are incorrect. It should be per m2.

No, W m-1 K-1 is correct.
 
Oops. Sorry, you're right. My mistake.

Chet
 
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