Heat Transfer Through a Styrofoam Container

[Erubus]

Homework Statement

A styrofoam cooler is in the shape of a rectangular box. Its interior dimensions are 30×40×60 cm. Its walls are 1.5 cm thick. It contains 4.0 kg of ice with the remaining space filled with water. The ice and water are in thermal equilibrium. The outside temperature is 25◦C, the coefficient of heat transfer for styrofoam is 0.023 W/m·K, and the latent heat of fusion for ice is 330kJ/kg. How long will it take for all the ice to melt?
(Answer: 8.9 hours)

Homework Equations

Q = L*Δm

$\frac{dQ}{dt}$ = $\frac{kA}{l}$*(T1- T2)

The Attempt at a Solution

Since the inside is a mixture of water and ice, the internal temperature is 0°C.

I use the latent heat equation because the inside is only going through a phase change.


I set the two equations equal to each other so that:
$\frac{kA}{l}$*(T$_{outside}$ - 0) = L*$\frac{dMass}{dt}$


and changing it so I can take an integral:
dM = $\frac{kA}{l*L}$(T$_{outside}$)dt


$\int^{0}_{4}$dM = $\frac{kA}{l*L}$(T$_{outside}$)$\int$dt

but the integral doesn't work out because integrating from the initial mass of the ice (4kg) to the final (0kg) ends up as ln(-4)

I think I set up the two equations incorrectly at the start.

 

[Chestermiller]

No. You did pretty well at setting it up. But where did the natural log come from? The integral of dM is M.

Also, there is a problem with terminology in the problem statement. 0.023 in this context should not be referred to as the "coefficient of heat transfer." It should be referred to as the thermal conductivity. In heat transfer parlance, the "coefficient of heat transfer" has an entirely different meaning.