capacity of the drink is 4.18 J/g·°C? (Assume that the heat transfer is 100% efficient.)

Answer 1

Your question lack clarity.

Answer 2

oops here.

When ice melts, it absorbs 0.33 kJ per gram. How much ice is required to cool a 12.0 fluid ounce drink from 75°F to 35°F, if the heat capacity of the drink is 4.18 J/g·°C? (Assume that the heat transfer is 100% efficient.)

Answer 3

Convert 12 ounces to grams.
(mass ice x 330 J/g) + [(mass water x specific heat water x (Tfinal-Tinitial)] = 0
Solve for mass ice.

Answer 4

thanks. um. i got 201. but it said it was wrong

Answer 5

how do i know what that mass of the ice is?

Answer 6

oops. forget what I said last

Answer 7

I got it wrong. can u explain it better. i don't think that is in the correct equation form??