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 0.016m^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？
Can anyone please give me some hints to do this?Thanks a lot!
The deltaTemp is (35-0)C.
Q=Area/Thickness*Heatconductivity*deltaTemp
For styrofoam, assume about .03Joules/(sec*m*C)
Solve for q,the amount of heat.
Now, finally a= massice*Lf
Q=Area/Thickness*Heatconductivity*deltaTemp
How you get this equation?
and how you use the styrofoam=.03 J/s*m*C?
To determine the amount of ice that melts in one hour, you need to calculate the amount of thermal energy transferred from the outside surface of the cup to the ice, and then convert it to the mass of ice melted using the latent heat of fusion.
Here's a step-by-step guide on how to solve the problem:
1. Calculate the thermal energy transfer using the formula: Q = k * A * ΔT / d.
- Q is the thermal energy transfer (in joules).
- k is the thermal conductivity of Styrofoam cup (Styrofoam's thermal conductivity is around 0.045 W/(m·K)).
- A is the surface area of the cup (in square meters).
- ΔT is the temperature difference between the inside and outside surface of the cup (in kelvin or degree Celsius).
- d is the thickness of the cup (in meters).
2. Convert the thermal energy transfer (Q) to the mass of ice melted using the formula: Q = m * L.
- Q is the thermal energy transfer (in joules).
- m is the mass of ice melted (in kilograms).
- L is the latent heat of fusion for ice (3.35 * 10^5 J/kg).
3. Rearrange the formula from step 2 to solve for the mass of ice melted:
m = Q / L.
4. Convert the time from one hour to seconds, since the thermal energy transfer (Q) is measured in joules per second (Watts).
5. Plug in the given values into the formulas and calculate the mass of ice melted.
I hope this explanation helps you in solving the problem.