Recent content by KylieB

or to remove the energy is it simply 150.48 = 75 x 4.186 x Tf 150.48 = 315.95 (T?) T? = 150.48/313.95 T? = 0.483degrees C then remove this from the initial temp of water 10 - 0.483 = 9.517 Deg C is this right so far?

So here's what i think is right so far Raise the temp of ice to 0 18 x 2.09 x 4 = 150 joules Remove this energy from the water 150.48=75 x 4.186 x(10-Tf)is this part right? 150.48= 3139.5-313.95(Tf) I am not sure how to get to the resultant temp of the water propery before I consider a phase...

Homework Statement An 18g ice cube at -4.0ºC is placed into 75g of water at 10 ºC in an insulated container. a. What is the final temperature of the system? b. How much of the ice melts? I have a sheet full of these & ice to water to steam I really need somone to explain how to find the...