If you mixed 200gm of ice that is at -5 deg C with 20g of water that is at 15 de

Question

If you mixed 200gm of ice that is at -5 deg C with 20g of water that is at 15 deg C, what will be the temperature and condition of the final state once equilibrium is achieved.

I know its gonna be ice but my reading is wrong for some reason i am getting -76 deg Celsius :D
I did this:
Mi = 200g
Mw = 20g
Ti = -5 deg C
Tw = 15 deg C
Tf = ?
MwCw(Tw-Tf) = MiLf + MiCw (Tf - Ti)
20g * 1(cal/g) *(15 - Tf) = 200g*80 cal/g + 200*1cal/g(Tf +5)
....

Solving for Tf gives:
-16700 = 220Tf
Tf = -75.9 deg C

which is clearly implausible since it has to be between -15 and 5 deg C.

Thanks in advance

Explanation / Answer

Mi = 200g
Mw = 20g
Ti = -5 deg C
Tw = 15 deg C
Tf = ?
the internal energy stored in water=MwCw(Tw-Tf)

   =20*1*(15-0)

   =300 cal -----------------(1)

the internal energy stored in ice at -5 deg c =MiLf + MiCi (Tf - Ti)

   =200*80+200*0.5*(0-(-5))

   =16000+500

   =16500 cal ---------------(2)

by obsurving this energies

300 cal is very less either for 16000,or 500

so

it is more in ice only, so the temparature -5 deg c ,is decreased to someT. similarly water temparature also becomes T

according to first law of thermodynamics

300=200*0.5*(T-(-5))

T=-2 deg c

i hope it is understand by you..........GOOD LUCK

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