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.)

Question

please explain realy realy well

thanks:)

Answer 1

Here is what I posted previously.

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.

Show your work and I'll try to find the error. Mass ice is the only unknown. mass water is 12 ounces; you need to convert that to grams. Specific heat water you know as well as Tf and Ti. There is no explaining to do.

Answer 2

this is wut i got:

12oz=340.194 grams

(miX330j/g) + [(340.194gX4.184X40]=0

56934.86 + 330j/g(mi)=0

-56934.86=330j/g(mi)

-56934.86/330j/g = -172.5298=(mi)

Answer 3

oops sorry I am rebekah

Answer 4

my sis changed it 4 her

Answer 5

ok. The error is in the 40. It should be a -40. It works this way.
(mass ice x 330) + [(mass water x specific heat water x (Tfinal-Tinitial)] = 0

(330X) + [(340.2 x 4.184 x (35-75)] = 0
330X + [(340.2 x 4.184 x (-40)] = 0
I will let you finish but I think you can see that it changes only the sign so you end up with +172 g and not -172 g(the negative sign makes no sense, of course).

Answer 6

Some quick comments. The F must be converted to C, the ice, when it melts at zero C must also be heated to 35 F (converted to C), and the fluid ounce is not the same as the the ounce in mass units.

Answer 7

how much ice is required to cool a 10.0 oz drink from 80 f to 31 f, if the heat capacity of the drink is 4.18 j/gc?

Answer 8

This answer is fine because -40 F is equal to -40 C
I got the answer 172 g of ice needed.