An aluminum cube of mass 44.25 g, at 22.3 degrees celcius is immersed in a styro
ID: 2120984 • Letter: A
Question
An aluminum cube of mass 44.25 g, at 22.3 degrees celcius is immersed in a styrofoam cup filled with LN2 at -195.8 degrees celcius . After 120 seconds 34.50 g of LN2 evaporates due to the immersion of the cube. Calculate Lv of LN2 and the rate R (calories/second) at which heat from the cube evaporates LN2. (Note that the 34.50 g evaporates is due solely to the immersion of the cube and not by the heat transferring into the cup from the room.) Use the accepted value of c aluminum when calculating Lv. Use your calculated value of Lv when calculating the rate R. Thr accepted value of c aluminum is 900 J/kgK.
Explanation / Answer
mass of AL cube = 44.25 g
t= 22.3 deg
temp of LN2 = -198 deg ;
after immersion 34.5 g LN2 evaporates.
the 34.50 g evaporates is due solely to the immersion of the cube
by law of conservation of energy ;
heat lost by al cube = heat gained by LN2;
mcdt = ml ;
l = latent heat of vapourisation;
44.25 *900*(22.3+198) = 34.5 * l;
l=254,302 J/kgk;
rate of vapourisation = ml/t ;
=73,112 j/s
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