Thanks ahead of time. Exercise #20 Calorimetry On a 10.5 ounce bag of potato chi
ID: 546874 • Letter: T
Question
Thanks ahead of time.
Exercise #20 Calorimetry On a 10.5 ounce bag of potato chips, the manufacturer states which is equivalent to 120, Calories. How many kilojoules are in this bag of potato chips? 1. How much energy is required to heat 25.0 grams of aluminum from 0.00 °C to 25.00 °c? The 2. specific heat of aluminum is 0.897 & K To inflate a balloon you must do pressure-volume work on the surroundings. If you inflate a balloon from a volume of 0.100 L to 1.85 L against an external pressure of 1.00 atm, how much work (in joules) is done 3.Explanation / Answer
Ans. #1. Mass of potato chips = 10.5 oz
Calorific value of potato chips = 120 Cal/ oz
Now,
Total Cal content of 1 packet chips = Total mass of chips x Calorific value
= 10.5 oz x (120 Cal/ oz)
= 1260 Cal
# Note that 1 dietary Calorie is equal to 1000 calories – the difference is often cited as capitalization of C.
So,
1260 Cal = 1260 kilo calories ; [1 calorie = 4.184 J]
= 1260 x 4.184 kilo J
= 5271.84 kJ
#2.The amount of required energy is given by-
q = m s dT - equation 1
Where,
q = heat gained or lost
m = mass
s = specific heat
dT = Final temperature – Initial temperature
dT = 25.000C – 0.000C = 25.000C = 25.00 K
Note: Difference in temperature is the same in 0C and kelvin units.
Now,
Putting the value sin equation 1-
q = 25.0 g x (0.897 J g-1 K-1) x 25.00 K
Hence, q = 560.625 J
Hence, required amount of energy = q = 560.625 J
#3. Change in volume, dV = Final volume – Initial volume
= 1.85 L – 0.100L
= 1.75 L
Given, pressure, P = 1.00 atm
Now,
Work done during inflation, W = P (dV)
Or, W = 1.00 atm x 1.75 L
Or, W = 1.75 atm L ; [1 atm L = 101.33 J]
Or, W = 1.75 x (101.33 J)
Hence, W = 177.3275 J
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