7.1 A twelve pack of your favorite beverage (12 fl. oz. cans) is placed in a coo

Question

7.1 A twelve pack of your favorite beverage (12 fl. oz. cans) is placed in a cooler at 70o F. How much heat must be removed (Btu’s) to cool the beverages to 34o F?

7.2 Seven pounds of ice at 20o F is placed in the cooler with the beverages in problem 7.1. Assume no heat is gained from the environment outside of the cooler. What is the equilibrium temperature (oF) of the beverages and the ice?

7.3 If the rate of heat gained by the cooler (in problems 7.1 and 7.2) from the surrounding environment is 50.0 Btu/hr., how long (hr.) will the ice in the cooler last? Assume the beverages are placed in the cooler at 32oF.

7.4 A runner consumes 1.0 L of water during an event. What has a greater cooling influence on the human body, the cold temperature of the water (5o C), or evaporation of this water from the skin of the runner? Justify your answer with supporting calculations.

Explanation / Answer

Temperature drop = 70 - 34 = 36 F = 20 C

Mass = 12 fl. oz = 354.88 g

C for beverage = 4.2 J/g-K (assumed water)

So heat removed = mCT = 354.88*4.2*20 = 29809 J = 29809/1055 BTU = 28.26 BTU

7.2 Mass of ice = 7 poound = 3175.15 g

Temp = 20 F

Mass of beverage = 12 fl. oz = 354.88 g

temp = 70 F

Heat released to reach 32 F = 354.88 * 4.2 * (38/1.8) = 31466 J

Heat released on freezing it = 354.88*336 = 119239 J

Heat required to heat ice to 32 F = 3175.15*2.03*(12/1.8) = 42970 J

So we see that the heat released in freezing the beverage is more than heat required to heat ice to 32 F

But the heat reuired to heat ice to 32 F is more than heat released to bring beverage to 32 F in liquid form

So the whole system will remain at 32 F, with some portion liquid and some frozen

the equilibrium temperature (oF) of the beverages and the ice = 32 F

7.3 Heat required to heat ice to 32 F = 3175.15*2.03*(12/1.8) = 42970 J

Heat required to melt it = 3175.15*334 = 1060500 J

So total heat extracted to melt = 1103470 J = 1045.88 BTU

So time for which ice will last = 1045.88 / 50 = 20.92 hours

7.4

1 L water = 1000 g

Body temperature = 37 C

heat extracted from body as water heats to 37 C = 1000*4.2*(37-5) = 134400 J

Heat extracted on evaporation = mass* latent heat of evaporation = 1000* 2260 = 2260000 J

So we see a lot more heat is taken away due to evaporation than due to the cold temperature of water. SO evaporation has more influence.

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