Thermal Equilibrium?

Question

If 50.0 g of ice at 0.0 degrees C is addd to 150.0 g of water at 80.0 degrees C, what will the temperature of the mixture be when it reaches thermal equilibrium?

how do you solve that after you change the g into kg and the celcius into kelvin?

Answer 1

Q(before) = Q(after)

Q(ice melting) + Q(ice water warming up) + Q(hot water cooling down)=0

Cf- latent heat of fusion )=334 J./gm =Joules/gm
Cv - specific heat of water = 4.1813 Joules/(gm K)

Q(ice melting)=m1Cf= 50.0 x 334 =16700 joules
Q(ice water warming up)=m1Cv(T-0)
Q(hot water cooling down)=m2Cv(80-T)
Note that the difference between K and C is the same
now we have
m1Cf +m1Cv(-0) +m1Cv(T)+m2Cv(80)+m2Cv(-T)=0
m1Cv(T)+m2Cv(-T) = m1Cf +m1Cv(-0) +m2Cv(80)
T=[m1Cf +m2Cv(80)]/[m1 - m2]Cv
T=[16700 joules +150 x 4.1813 ]/ [(50 -150) x 4.1813 ] = -41C
(the negative sign represents that the temperature has dropped)

Actual temperature of the water will be 41 C.