At a picnic, a Styrofoam cup contains lemonade and ice at 0 degree C. The thickness of the cup is 2.0*10^-3m, and the area is 0.016 m^2. The temperature at the outside surface of the cup is 35 degree C. The latent heat of fusion for ice is 3.35*10^5 J/kg. What mass of ice melts in one hour?
I had posted this question several hours ago, and someone gave me such solution:
You need the heat transfer coefficient of styrofoam. It is about .01 watts meters/deg
call area of foam A = .016 m^2
call temperature difference across styrofoam = 35 - 0 = 35 deg C
call thickness of foam t = 2*10^-3 m
then heat gained through foam = (k A/t)((35-0)
heat gain rate watts = (.01*.016/.002)(35)
That watts * time in seconds = Joules
so that times 3600 = Joules gained to melt ice in one hour. Now kg melted = Joules gained/ heat of fusion in Joules/ kilogram
However, I don't quite get the last part, can someone please explain? Thanks!
we got the heat gain in watts
watts * time = Joules
(.01*.016/.002)(35) was our watts
so
(.01*.016/.002)(35) Joules/s * 3600 seconds = X * 3.35*10^5 Joules/kg
Solve for X kilograms of ice
The last part of the solution is calculating the mass of ice melted by using the heat gained through the foam and the latent heat of fusion for ice.
To understand this calculation, let's break it down step by step:
1. The heat gained through the foam is calculated using the formula:
heat gained = (k * A/t) * (35 - 0)
Here, k is the heat transfer coefficient of the Styrofoam cup (0.01 W m/deg), A is the area of the cup (0.016 m^2), and t is the thickness of the cup (2.0 * 10^-3 m). The temperature difference across the cup is 35 - 0 = 35 deg C.
2. The heat gain rate is calculated by dividing the heat gained by the thickness of the cup:
heat gain rate (in watts) = (0.01 * 0.016 / 0.002) * 35
This gives us the rate at which heat is gained through the cup in watts.
3. To determine the total energy gained in one hour, we need to convert the heat gain rate from watts to joules. We multiply the heat gain rate by 3600 (the number of seconds in one hour):
energy gained (in joules) = heat gain rate (in watts) * 3600
This gives us the total amount of energy gained by the cup in one hour.
4. Finally, to calculate the mass of ice melted, we divide the energy gained by the latent heat of fusion for ice (3.35 * 10^5 J/kg):
mass of ice melted = energy gained (in joules) / latent heat of fusion (in J/kg)
Dividing the energy gained by the latent heat of fusion gives us the mass of ice that can be melted by that amount of energy.
Hope this explanation helps clarify the last part of the solution!