A 25 g glass tumbler contains 350 mL of water at 24°C. If four 20 g ice cubes each at a temperature of -3°C are dropped into the tumbler, what is the final temperature of the drink? Neglect thermal conduction between the tumbler and the room.
Q = mcT
Q = mLf (energy required to melt a substance of mass m)
The Attempt at a Solution
Here are all the known variables:
massw = .35kg
c (specific heat for water) = 4186 J/kg * C
Twater,initial = 24 C
massice = 4*20g = .08kg
Lf = 3.33*105 J/kg
Tice,initial = -3 C
Tf = ?
First I can use Q = mcT to see if all the ice will get melted by the water
Q = (.35kg)*(4186 J/kgC)*(24C) = 35162.4 J
m = Q/Lf = (35162.4 J) / (3.33*105 J/kg) = .105592
Since the total mass of the ice is .08, then all the ice will get melted.
Now we can apply conservation of energy to find the final temperature.
massw*c*(Twater,initial - Tf) = massice*Lf + massice*c*(Tf-Tice,initial)
Plugging in all my variables and solving I get 4.18 C which is incorrect.
My math could be wrong but I don't believe it is.
Could anyone provide any insight on what I am doing wrong?