You are thinking ahead to when one of your friends is having an outdoor wedding.

Question

You are thinking ahead to when one of your friends is having an outdoor wedding. Your plan is to design the perfect lemonade for the event. The problem with lemonade is that you make it at room temperature and then add ice to cool it to a pleasant 10 degrees C. Usually, the ice melts diluting the lemonade too much. To help you solve this problem, you look up the specific heat capacity of water (1.0 cal/(gmC)), the specific heat capacity of ice (.50 cal/(gmC)), and the latent heat of fusion of water (80 cal/gm). You assume that the specific heat capacity of the lemonade is the same as water. Since you will cool your lemonade in a thermos jug, assume no heat is added to the lemonade from the environment. Using that information, you calculate how much water you get from all the ice melting if you make 6 quarts (5.6 kg) of lemonade at room temperature (23 degrees C) and add ice which comes straight from the freezer at -5 degrees C.

Explanation / Answer

So first, we need to know that H = m*s*T

In this case, since no heat is added to the lemonade from the environment, we know that all the heat released by the lemonade is absorbed by the ice. Thus, overall H = 0

Hrelease by lemonade + Habsorb by ice = 0

Note that when heat is released, we consider Hrelease < 0

M = mass of lemonade = 5.6 * 103 kg = 5600 gm

S = specific heat capacity of water = 1.0 cal/(gmC)

m = mass of ice melted, this is the quantity that we are finding.

s = specific heat capacity of ice = 0.5 cal/(gmC)

sf = latent heat of fusion of water = 80 cal/gm

Hrelease by lemonade + Habsorb by ice = 0

[M*S*T] + [m*s*T + m * sf ] = 0

[5600*1*(10 - 23)] + [m*0.5*(0 - (-5)) + m*80 + m*1*(10-0)] = 0

m * (2.5+80+10) = 5600*(13)

m = 5600*(13)/(2.5+80+10) = 72800/92.5 = 787.03 gm

You will need to melt about 787 gm of ice to cool the lemonade to 10 degree C.

When 787 gm of ice are all melted, there would be 787 gm of water added to the original lemonade.

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