How long does it take the sun to melt a block of ice?

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SUMMARY

The discussion focuses on calculating the time it takes for the Sun to melt a block of ice at 0°C with a specified area and thickness. The relevant equation used is Q/t = IeAcos(theta), where Q is the heat energy, t is time, Ie is the solar irradiance, A is the area, and theta is the angle of incidence. The original poster incorrectly manipulated the equation, resulting in an incorrect expression for time. The correct approach requires careful handling of the variables to derive the accurate time for melting the ice.

PREREQUISITES
  • Understanding of thermodynamics, specifically heat transfer principles.
  • Familiarity with solar irradiance calculations.
  • Knowledge of the properties of ice, including density and latent heat of fusion.
  • Proficiency in trigonometric functions, particularly cosine for angle calculations.
NEXT STEPS
  • Review the principles of heat transfer and the specific heat equation.
  • Learn about solar irradiance and its impact on energy calculations.
  • Study the properties of ice, including density and latent heat of fusion.
  • Practice manipulating equations involving multiple variables to avoid common errors.
USEFUL FOR

Students studying physics, particularly those focusing on thermodynamics and energy transfer, as well as educators seeking to clarify concepts related to solar energy and phase changes in materials.

angelbluw15
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Homework Statement


How long does it take the Sun to melt a block of ice at 0°C with an area of 1.10 m2 and thickness 1.07 cm? Assume that the Sun's rays make an angle of 31.1° with the normal to the area, and that the emissivity of ice is 0.0400.


Homework Equations


Q/t = IeAcos(theta)



The Attempt at a Solution


ok, I've done multiple attempts and am still not getting the correct answer!

1) Q/t = IeAcos(Theta)
Q/t = (1000 W/m^2)(0.04)(1.10cos(31.1)) = 37.6 J/s
but from this number, I don't understand how to find time, should it be one??

or
A*h *ρ *L /t = IeACOSθ
t = time for ice to melt = IeCOSθ /h *ρ *L
t = (1000 W/m^2)(0.04)(cos(31.1)/(0.0107 m)(917 kg/m^3)(3.33*10^5 J/kg)
t= 1.05*10^-5 s

but, that doesn't work either. I've been trying this problem all afternoon, please help!
 
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angelbluw15 said:
A*h *ρ *L /t = IeACOSθ
t = time for ice to melt = IeCOSθ /h *ρ *L
t = (1000 W/m^2)(0.04)(cos(31.1)/(0.0107 m)(917 kg/m^3)(3.33*10^5 J/kg)
The OP has blundered in manipulating the equation, yielding, numerically, 1/t instead of t.
 

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