Calculating Time to Melt 8.78kg of Ice in a Styrofoam Icebox

[espnaddict014]

Approximately how long should it take 8.78 kg of ice at 0oC to melt when it is placed in a carefully sealed Styrofoam icebox of dimensions 27.2 cm x 35.7 cm x 51.2 cm whose walls are 1.20 cm thick? Assume that the conductivity of Styrofoam is double that of air and that the outside temperature is 30.9oC.


I'm trying to start with the equation:

delta Q/ delta t = kA* [(T1-T2)/l]

But I have a volume, and I guess I didn't know if V could replace A, but it doesn't seem like it should. Then even if it could, I would have calculated a J/s. And also, do i need to incorporate a phase change in this problem? Any help would be appreciated.

 

[Andrew Mason]

Approximately how long should it take 8.78 kg of ice at 0oC to melt when it is placed in a carefully sealed Styrofoam icebox of dimensions 27.2 cm x 35.7 cm x 51.2 cm whose walls are 1.20 cm thick? Assume that the conductivity of Styrofoam is double that of air and that the outside temperature is 30.9oC.


I'm trying to start with the equation:

delta Q/ delta t = kA* [(T1-T2)/l]

But I have a volume, and I guess I didn't know if V could replace A, but it doesn't seem like it should. Then even if it could, I would have calculated a J/s. And also, do i need to incorporate a phase change in this problem? Any help would be appreciated.

Work out the total surface area of the icebox. That is A. You are given the thermal conductivity, k and the thickness, l. Just plug in the numbers to give the rate of heat transfer. You should be able to work out the length of time needed to melt the ice from that.

AM

 

[thepatientmental]

To calculate the time it takes for 8.78 kg of ice to melt, we can use the formula for heat transfer:

Q = m * L

Where Q is the heat required to melt the ice, m is the mass of ice, and L is the latent heat of fusion for water which is 334 kJ/kg.

First, we need to convert the dimensions of the icebox from cm to m:

27.2 cm = 0.272 m
35.7 cm = 0.357 m
51.2 cm = 0.512 m

Next, we can calculate the volume of the icebox:

V = l * w * h = (0.272 m)(0.357 m)(0.512 m) = 0.05 m^3

Since the walls of the icebox are 1.20 cm thick, we need to subtract this from the dimensions to get the volume of the actual space inside the icebox:

V' = (0.272 m - 0.024 m)(0.357 m - 0.024 m)(0.512 m - 0.024 m) = 0.04 m^3

Now, we can calculate the volume of ice:

Vice = 8.78 kg / 917 kg/m^3 = 0.0096 m^3

The remaining volume is filled with air, so we can calculate the mass of air:

mair = (0.04 m^3 - 0.0096 m^3) * 1.2 kg/m^3 = 0.039 kg

Next, we can calculate the heat required to melt the ice:

Q = (8.78 kg)(334 kJ/kg) = 2934.52 kJ

Now, we can use the formula for heat transfer to calculate the time it takes for the ice to melt:

Q/Δt = k*A*(T1-T2)/l

Where:
Q = heat required to melt the ice (2934.52 kJ)
k = thermal conductivity (double that of air, so we will use 0.04 W/mK)
A = surface area of the icebox (0.04 m^2)
T1 = outside temperature (30.9°C = 304.05 K)
T2 = initial temperature of ice (0°C = 273.15 K)
l = thickness