Please show all works corretly. Thanks A 12 ounce can of a certain citrus-flavor

Question

Please show all works corretly. Thanks

A 12 ounce can of a certain citrus-flavored carbonated beverage contains 170 calories (A food "calorie" is actually a kilo-calorie). (a) How many Joules of energy is this equivalent to? (b) If an equal amount of energy was used to increase the can's gravitational potential energy (by lifting it straight up), how high would the can be lifted? The mass of the liquid and can combined is 390 grams, (c) If 170 kilocalories of heat were added to an initially "ice cold" beverage, what would the final condition of the soda be? Treat the liquid itself as if it were 355 grams of pure water, ignore the can, and find the final temperature.

Explanation / Answer

energy E = 170 kilo-cal

1 kilo-cal = 4184 J

a)

170 kilo-cal = 170 * 4184 J

711280 J

b)

m*g*h = 711280

390 * 10^-3 *9.8 * h = 711280

h = 186.1 km

c)

170 kilo-cal = 711280 J

this energy is used to raise the temp

m * c * (T) = 711280       [for water specific heat C = 4186 J/kg-c]

355 * 10^-3 * 4186 * T = 711280

T = 478.64

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