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 Thermodynamics and Fluid Mechanics Thermodynamics and Fluid Mechanics Physics Help Forum Nov 22nd 2011, 07:13 PM #1 Member   Join Date: Nov 2010 Posts: 33 Heat Transfer Equation for Ice Cube Melting Time The Heat Transfer Coefficient : h = q / A * delta T en.wikipedia..org/wiki/Heat_transfer_coefficient. Can be applied to phase changes between a solid and surrounding fluid. h = cal/sec/Area* Temp difference q = heat flow cal/sec A =M^2 T = temp difference So q = h * A * delta T I placed a 100 gram ice cube in one liter of water at 30 deg C = 30000 cal. 1 g. ice absorbs 80 calories melting so the cube absorbed 8000 calories from the water . So 22000 cal/ 1100 grams water = 20 deg final water temp. Thats the easy part. The ice took 6 minutes to melt. h for water = .14328 cal/sec/M^2 * T Area for ice cube = .01224 M^2 ( surface area of 100 cc cylinder) . delta T = 30 deg The problem is that this heat transfer equation did not produce close to 6 minutes : q = .143328 cal/sec/M^2* T ( .01224 M^2 ) ( 30 deg) = .0526 cal/sec and the melting time : 8000 cal/.0526 cal/sec is much greater than 6 min So what is the correct equation for this example ? Last edited by morrobay; Nov 22nd 2011 at 11:14 PM.   Nov 24th 2011, 12:02 AM   #2
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Join Date: Nov 2010
Posts: 33
 Originally Posted by morrobay The Heat Transfer Coefficient : h = q / A * delta T en.wikipedia..org/wiki/Heat_transfer_coefficient. Can be applied to phase changes between a solid and surrounding fluid. h = cal/sec/Area* Temp difference q = heat flow cal/sec A =M^2 T = temp difference So q = h * A * delta T I placed a 100 gram ice cube in one liter of water at 30 deg C = 30000 cal. 1 g. ice absorbs 80 calories melting so the cube absorbed 8000 calories from the water . So 22000 cal/ 1100 grams water = 20 deg final water temp. Thats the easy part. The ice took 6 minutes to melt. h for water = .14328 cal/sec/M^2 * T Area for ice cube = .01224 M^2 ( surface area of 100 cc cylinder) . delta T = 30 deg The problem is that this heat transfer equation did not produce close to 6 minutes : q = .143328 cal/sec/M^2* T ( .01224 M^2 ) ( 30 deg) = .0526 cal/sec and the melting time : 8000 cal/.0526 cal/sec is much greater than 6 min So what is the correct equation for this example ?
Thermal conductivity: H = delta Q/delta t = k A delta T/x
This equation gave a melting time of 5 min ,in agreement with data
k = for water. .6 Watt / M * T = .14328 cal/sec/M*T
Area= same above, .01224 M^2
delta T = 30 deg.
x = thickness of boundary layer , ice-water. = .002 Meter
( Journal Phys. Oceanography)

So : Q/t = .14328 cal/sec/M*T (30 deg) ( .01224 M^2) / .002 M = 26cal/sec
8000 cal/ 26cal/sec =307 sec = 5.1 min  Tags cube, equation, heat, ice, melting, time, transfer Search tags for this page
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# equation for ice melting

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