Question

A large punch bowl holds 3.35 kg of lemonade (which is essentially water) at 20.0° C. A 1.62-kg ice cube at
−10.2°
C is placed in the lemonade. What is the final temperature of the system, and the amount of ice (if any) remaining? Ignore any heat exchange with the bowl or the surroundings.
Answer must be in:
______degrees C
______ kg
For kg, the answer is not zero
Please please show all work ( even the algebra)
=)

Answer 1

energy will be xchanged

M1C(T3-T1) = M2C(T2-T1)

Answer 2

I keep trying to use that setup and I keep getting the wrong answers :(

Answer 3

well C gets cancelled

3.35 (20-T) = 1.623 (T+10.2)

Answer 4

I solved for T and got 10.14 and tried to enter it to my hw and it said it was wrong :/

Answer 5

well no idea then, sorry

Answer 6

Thank you for trying

Answer 7

add latent heat of fusion

Answer 8

and google for sp. heat capacity of ice

Answer 9

I found that latent heat of fusion for water/lemonade is 33.5 x 10^4, Specific heat of water is 4186, and sp. heat for ice is 2090...but I don't know where to put them in the equation

Answer 10

what's the answer 0 C??

Answer 11

that's why I am having so much trouble with it bc there no example like it, not even in the book, I haven't tried 0 yet i will right now but I have no idea what to input for kg

Answer 12

so I will try putting 0 degree C, but do you know how I can figure out the mass of ice ?

Answer 13

http://www.wolframalpha.com/input/?i=1.62+-+%283.35*4200*%2820%29+-+1.62*2108*10.2%29%2F334000

Answer 14

It's correct! Thank you so so so much!!!!

Answer 15

i believe you understood what i did .
you are welcome

Answer 16

yes I do, I will rework it again just in case! Thank you again =)

Answer 17

just think what will happen ... (first the ice will come at 0 C) ...
still there is some heat remaining in the water ... that heat will melt ice.

Answer 18

I think I got more caught up in the math than the concept, thank you for clarifying everything!