Physics Help for this Problem!?

Question

A 2-liter pitcher of lemonade has been sitting on the picnic table in the sun all day at 33 degrees Celsius. You pour 0.24 kg into a Styrofoam cup and add six ice cubes (each 25 grams at the melting point). Assuming no heat is lost to the surroundings, what is the final temperature and compostition of the mixture in the cup? Assume specific heat for lemonade is the same as water

Answer 1

Ok, you pour 0,24 kg into a glass, sorry, let me call it a glass, so it will be : 240 grams

So you must use this :

the heat that the water loses, will be gain by the ice, right ?

If the ice is at the meltin temperature, so :

240*1*(33-T) = 6*25*0.5*T

7920 - 240t = 75T

then T = 25.39 ºCelsius >>> final temperature

The ice is at the meltin point, so you dont need to use the change from ice to water, because the ice is already at 0ºCelsius

Hope that might help you

Answer 2

The temperature of lemonade is also=33 degrees
Now 0.150 kg of ice is introduced.
In problems of this kind it is better to find out the amount of heat that should be lost by lemaonade to attain 0degree temperature.
Then find the massof ice that has to be fused to water at 0degree to produce energy equivalent tothat obtained in the earlier case.
If mass of ice required is less than the actual amount you have
then the lemonade and remaining ice and the mass of fused ice which now exists as water at 0degree ,remain in equilibrium at
melting point ofice
Here heat required to be lost by lemonade to attain 0degree
is=0.24*4186*33=33153.12J
Mass of heat that has tobe fused to water at 0degree to compensate this amount of heat=33153.12/latent heat of fusion
=33153.12/(3.36*10^5)=0.09867kg
This is lesser than amount of ice supplied
So 0.15-0.09867=0.05133kg of ice remins as it is.
Therefore 0.098kg water at 0degree , 0.0513kg ice at degree and lemonade 0.24 kg existin equilibrium at 0 degree

Answer 3

come on, do your homework yourself...