Recent content by jiceo1

[2220 J/kgC * 0.720kg * 0-(-10)] + [0.720kg * 333000 J/kg * 4190J/kgC * 0.720kg * (T-0)] = 4190 J/kgC * 2.5kg * (25-T) T=1.5 degrees

for part B, since energy gained = energy lost. Energy gained = the ice warming up to 0, then melting, then the water at 0 degrees warming up to a final temperature. Energy lost = water cooling down to a final unknown temperature. I got 1.5 degrees for the final temperature. Correct?

ooh, I see it... as the water warms up, it warms from 0 degrees to 15 degrees. -_- my bad... so my final answer is: 300,996 J

oh, ok: so for part A it would be: (2220 J//kgC)(0.720kg)(0-(-10)) + (0.720)(333000 J/kg) + (4190 J/kgC)(0.720kg)(15-(-10)) = 331000 J correct?

I have another related question, so I'm going to post it here instead of creating a new topic: An ice cube has a mass of 720 g and a temperature of -10° C. Assume that the specific heat of ice is c_ice = 2220 J/kg•°C, the latent heat of fusion of ice is L_F = 333 KJ/kg and the specific heat of...

I assume that since half of the ice is still there in the water, then the energy gained from the hot plate has all gone to melting half of the ice's mass. So at this point, the temperature would still be 0 degrees, am I wrong?

Homework Statement A beaker containing 50g of ice and 250g of liquid water is initially at 0celsius. It is then heated on a hot plate until half of the ice is melted. What is the temperature of the ice-water mixture at that point? A. 2.0 C B. 0.5 C C. 1.0 C D. 1.5 C E. 0.0 C Homework...