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Old Nov 22nd 2011, 08:13 PM   #1
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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 23rd 2011 at 12:14 AM.
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Old Nov 24th 2011, 01:02 AM   #2
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Originally Posted by morrobay View Post
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
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