Question

A large punch bowl holds 3.90 kg of lemonade (which is essentially water) at 25.0 ∘C. A 5.90×10−2-kg ice cube at -11.0 ∘C is placed in the lemonade. 1. What is the final temperature of the system? Ignore any heat exchange with the bowl or the surroundings. 2.What is the amount of ice (if any) remaining?

Answer 1

To bring the punch to zero would take Q = mc(delta T) Joules

Q = (3.90)(4186)(25) = 408135 J

For the ice to get to zero, Q = (.059)(2090)(11) = 1356.41 J

For the ice to melt Q = mH = (.059)(3.34 X 10⁵) = 19706 J

Total = 21062.41 J

Since that is less than the need to bring the punch to zero, all of the ice melts, that answers part B

For part A, then

mc(delta T) + mH + mc(delta T) = mc(delta T)

(.059)(2090)(11) + (.059)(3.34 X 10⁵) + (.059)(4186)(Tf) = (3.9)(4186)(25 - Tf)

21062.41 + 246.974Tf = 408135 - 16325.4Tf

16572.374Tf = 387072.59

Tf = 23.356 ^o C <<<<<<<answer part A

for part B , there will be no ice left all will melt.