Asked by Serena
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!
Can anyone please give me some hints to do this?Thanks a lot!
Answers
Answered by
bobpursley
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
For styrofoam, assume about .03Joules/(sec*m*C)
Solve for q,the amount of heat.
Now, finally a= massice*Lf
Answered by
Serena
Q=Area/Thickness*Heatconductivity*deltaTemp
How you get this equation?
and how you use the styrofoam=.03 J/s*m*C?
How you get this equation?
and how you use the styrofoam=.03 J/s*m*C?
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