It is a hot summer day and Chris wants a glass of lemonade. There is none in the

Question

It is a hot summer day and Chris wants a glass of lemonade. There is none in the refrigerator, so a new batch is prepared from freshly squeezed lemons. When finished, there are 208 grams of lemonade at 27.0°C. That is not a very refreshing temperature, so it must be cooled with ice. But Chris doesn’t like ice in lemonade! Therefore, just enough ice is used to cool the lemonade to 10.2°C. Of course, the ice will melt and reach the same temperature. If the ice starts at -11.9°C, and if the specific heat of lemonade is the same as that of water, how many grams of ice does Chris use? Assume there is no heat transfer to or from the surroundings.

Data for water at 1 atm: Melting point = 0.0°C Specific heat liquid = 4.18 J/g·°C Specific heat solid = 2.06 J/g·°C Heat of fusion = 333 J/g It is a hot summer day and Chris wants a glass of lemonade. There is none in the refrigerator, so a new batch is prepared from freshly squeezed lemons. When finished, there are 208 grams of lemonade at 27.0°C. That is not a very refreshing temperature, so it must be cooled with ice. But Chris doesn't like ice in lemonade! There fore, just enough ice is used to cool the lemonade to 10.2°C. Of course, the ice will melt and reach the same temperature. If the ice starts at-11.9°C, and if the specific heat of lemonade is the same as that of water, how many grams of ice does Chris use? Assume there is no heat transfer to or from the surroundings Data for water at 1 atm: Mclting point-0.0°c Specific heat liquid 4.18 Jg C Specific heat solid 2.06 J/g. C Heat of fusion-333 J/g

Explanation / Answer

Heat lost by lemonade = heat gained by ice

heat lost by lemonade = mCpdT

m = mass, Cp = specific heat, dT = change in temperature

heat gained by ice = heat gained to melt ice + heat to raise temperature of water

So,

with x g ice taken,

heat lost by lemonade = 208 x 4.18 x (10.2 - 27) = -14606.6 J

heat gained by ice = x g x 333 + x g x 4.18 x (10.2 + 11.9) = 14606.6

solving for x,

x = 34.34 g

So mass of ice added = 34.34 g

           

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