Asked by Om
A peice of ice weighting 5g at -20 centigrade is put into 10g of ice at 30C. Assuming no heat is lost to the outside.Calculate the final tempature of the mixture?
Answers
Answered by
Damon
Ice at 30 deg C??? Not likely but anyway
specific heat of ice (about 2 J/gram deg c) does not matter in the end here
call it sh
heat into cold ice = heat out of hot ice
5 (T - -20) * sh = 10 (30-T) * sh
5 T + 100 = 300 - 10 T
15 T = 300
T = 20
BUT I doubt the whole thing
suspect that 10 grams was water at 30 deg
In that case
Heat in to heat the ice up to zero
5 (20)*sh
Heat in to melt the ice
5 * hm where hm is the heat of fusion of water
Heat in to warm melted ice up to T
5(T-0) * shw
where shw is specific heat of water
Heat out of 30 deg water
10(30-T)*shw
so
5 (20)*sh + 5 * hm + 5(T-0) * shw = 10(30-T)*shw
If T comes out below zero, then T = 0 and some of the 30 deg water froze
specific heat of ice (about 2 J/gram deg c) does not matter in the end here
call it sh
heat into cold ice = heat out of hot ice
5 (T - -20) * sh = 10 (30-T) * sh
5 T + 100 = 300 - 10 T
15 T = 300
T = 20
BUT I doubt the whole thing
suspect that 10 grams was water at 30 deg
In that case
Heat in to heat the ice up to zero
5 (20)*sh
Heat in to melt the ice
5 * hm where hm is the heat of fusion of water
Heat in to warm melted ice up to T
5(T-0) * shw
where shw is specific heat of water
Heat out of 30 deg water
10(30-T)*shw
so
5 (20)*sh + 5 * hm + 5(T-0) * shw = 10(30-T)*shw
If T comes out below zero, then T = 0 and some of the 30 deg water froze
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