A large punch bowl holds 3.65 kg of lemonade (which is essentially water) at 20.
ID: 1591447 • Letter: A
Question
A large punch bowl holds 3.65 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. Please answer what's the final temprature and what is the remaining of the ice as well! there are 2 parts not only one. someone answered for the temperature as C but that was wrong. This is the second time I posted this I need help as soon as possible!
Explanation / Answer
Here,
mass of ice , mi = 1.62 Kg
mass of lamonde , mw = 3.65 Kg
heat needed to melt ice = mass * (Sice *(0 -(-10.2)) + latent heat of ice)
heat needed to melt ice = 1.62 *(2108 * 10.2 + 334 *10^3)
heat needed to melt ice = 5.76 *10^5 J
heat gain by water to reach 0 dgeree C
heat released = 3.65 * 4186 * 20
heat released = 3.06 *10^5 J
as the heat released is less than needed to melt the ice
the final temperature will be 0 degree C
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let the remaining ice is m
energy gain by ice = energy loss by water
(1.62 - m) * 334 *10^3 + 1.62 *2108 * 10.2 = 4186 * 20 * 3.65
solving for m
m =0.81 Kg
the mass of ice remaining is 0.81 Kg
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