Question

What will be the temperature of the maximum if 100 gm ice at 0oc is put in 100 gm water at 80oc?

Answer 1

Given that
latent heat of fusion = 80 cal/g
sp. heat capacity of water =1 cal/g/°c
sp. heat capacity of ice =0.5 cal/g/°c

So first thing in this mixing process will occur, is heat transfer from water at 80°c to saturated ice at 0°c till thermodynamic equilibrium occurs.
Since ice is saturated, so it will absorb heat in the form of latent heat to convert into water.
Let final temp.( equilibrium) of mixture is = T°c
So energy balance equation,
Latent heat to ice at 0°c to convert into saturated water at 0°c + sensible heating of this saturated water from 0°c to final temp T°c = cooling of hot water from 80°c to final temp T°c

=> (540*80) + {540*1*(T-0)} = {540*1*(80-T)}

=> 540*T = -540*T

=> 1080*T = 0

=> T = 0

Answer 2

Answer:

HEY MATE HERE IS YOUR ANSWER !!!

Explanation:

Let t is final temperature of the mixture then heat taken by

Let t is final temperature of the mixture then heat taken by100gm of ice to melt =

=mL = 100*80

00*80= 8000 cal heat taken by water of 0 degree to raise it's temperature to t degree = mst = 1 00*1*t

00*1*t=. 100t therefore

100t thereforetotal heat required by ice

100t thereforetotal heat required by ice= 8000 + 100t

00theat given by 100gm of water at 80 degrees

80-t) = 100. (. 80-t)

80-t)heat taken=heat given

80-t)heat taken=heat given8000 + 100t=100(80-t)

80-t)heat taken=heat given8000 + 100t=100(80-t)8000+100t=8000–100t

80-t)heat taken=heat given8000 + 100t=100(80-t)8000+100t=8000–100t200 t=0

t=0t=0 dergee Celsius

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