An 8 cm3 ice cube (temperature = 0.00 °C) is dropped into a glass with 3 dL juice which results in the ice cube being melted. By doing this, the juice drops to a temperature of 3.00 °C. What was the temperature of the juice before the ice cube was added? (Treat juice as water in this assignment. The density for ice is 920
kg/m3. Assume no energy is lost to the surroundings).
m1=(920)(8E-6) = 7.36E-3
m2=(1000)(0.3) = 0.3kg
The Attempt at a Solution
Energy gained by ice = Energy lost by water
(Ice → Water0C) + (Water0C → Water3C) = Waterinitial → Water3C
m1Lf+m1c(T3-T0) = m2c(Tinitial-T3)
I am not sure whether in the case of water ΔT is (Tfinal-Tinitial) or vice versa.
I get either a negative temperature, which is bs, or after swapping them around Tinitial= 5°C which I do not quite believe, because who would want to cool down water by 2°C?!
Please help, I am on the verge of questioning my sanity after spending more than an hour on this...
(One year ago it would've been no problem for me, but after a year of little to no physics my brain went ..dumb :D)