Heat transfer question from surroundings to a thin sheet of ice

AI Thread Summary
The problem involves calculating the mass of ice that melts due to heat transfer from the surrounding air to a thin sheet of ice. The net radiant energy transfer to the ice is given as 20W, leading to a total energy of 1200J over one minute. The latent heat of fusion for ice is 333kJ/kg, which is essential for determining the mass of ice melted. Attempts to solve the problem yielded incorrect mass values, suggesting a misunderstanding of the heat transfer equations or the role of the provided heat transfer coefficient. Clarification is sought on the necessity of the heat transfer coefficient (q) in solving the problem.
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Homework Statement



A thin sheet of ice with an area of 0.15m^2 on each side hangs from a roof.
The ice is at 0C (Celsius) and the surrounding air is at 10C.
The net rate at which radiant energy is transferred to the ice from the sunlight and
from the surroundings is 20W; q=9.5W*m^-2*K^-1
What mass of ice melts in 1 minute?

Homework Equations


Heat trasfer by convection: H=qAΔT (A: surface area)
Latent heat for phase change: ΔQ=Lm (L:latent heat, specific to material, m=mass)

The Attempt at a Solution


I'm in trouble. The correct answer is 8.74X10^-3kg but me and all of my classmates failed to get the correct answer. And I don't know why q is provided in the question. Do I really need that to solve this problem?
Anyways, this is my attempt to solve the problem.

The net rate at which radiant energy is transferred to the ice = 20W (J/s)
Thus, in 1 minute = 20Wx60sec=1200J=The heat ΔQ provided to change the phase of ice.
(solid to liquid)

Latent heat of fusion (water) = 333KJ/kg required (solid to liquid)
Therefore, 1200J=(333Kj/kg)*m
m=0.003604kg

or this:
the heat transferred from surrounding area to the sheet of ice:
H = qAΔT = (9.5W*m^-2*K^-1)(0.3m^2)(10k)=14.25W
in 1 minute = 14.25Wx60sec=855J
Latent heat of fusion (water) = 333KJ/kg (solid to liquid)
Therefore, 855J=(333Kj/kg)*m
m=0.002567kg

please tell me what I did wrong. Thanks a lot and have a nice day :D
 
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That's odd. Are you given the specific heat of the surrounding environment?
 
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