Question

A large punch bowl holds 3.45 kg of lemonade (which is essentially water) at 20.0° C. A 0.33-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 1

The amount of heat required to change the temperature of the ice from -10.2⁰C to 0⁰C is
Q = m c T
Q = 0.33 kg x 4186 x 10.2
Q = 14090.076 J
This heat is provided by the water at 20⁰C . The final temperature of the water after liberating this amount of heat
mc T = Q = 14090.1 J
T =  14090.1 J / 4186 x 3.45
20 - T = 0.976
T = 19.024⁰C
Now there is 0.33 kg of ice at temperature 0⁰C and 3.45 kg of water at 19.024⁰c
The amount of heat required by the ice to completely melt is
Q = m L
Where L is the latent heat of fusion
Q = 0.33 kg x 334000
Q = 110220 J
This heat is given by the water ,The final temperature is
Q = m C T = 110220 J
T = 110220 / 3.45 x 4186
19.024 - T =7.63
T = 11.394⁰C
The ice is completely melt ,now there is 3.45 kg of water at 11.394⁰C and 0.33 kg of water at 0⁰C
The final temperature is
M c T = m cT
3.45 x (11.394 - T) = 0.33 x (T- 0)
3.78 T = 39.3093
T = 10.4⁰C
The final temperature of the system is 10.4⁰ C, and there is no ice remaining