A 12 V lead-acid car battery, engineered for \'up to 500 or more charge/discharg
ID: 1436595 • Letter: A
Question
A 12 V lead-acid car battery, engineered for 'up to 500 or more charge/discharge cycles', has a rating of 135.0 A hr. (It sells for $144.00.) Calculate the total amount of charge that moves through the battery before it needs to be recharged.
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b) What is the total electrical energy that the battery can deliver before it needs to be recharged?
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c) The total mass of the battery is 18.1 kg. (It weighs 40 lb). What fraction of that mass is converted to electrical energy by the battery before it needs to be recharged?
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d) How many kilograms of such batteries would be required to get the same mechanical energy (essentially mileage, for cars of the same total weight) as 14 gallons of gasoline? Assume that the efficiency of an electrically powered car is 2.5 times that of a gasoline-powered car for the conversion to mechanical energy. DATA: Energy content of 1 gallon of gas = 1.32 x 108 J.
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e) One gallon of gasoline (mass = 2.8 kg) requires about 8.9 kg of Oxygen for combustion. What fraction of the mass is converted to energy?
Explanation / Answer
(a)
Here,
potential , V = 12 v
total capacity , E = 135 A.hr
total charge moved = 135 A * 3600 s
total charge moved = 4.86 *10^5 C
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(b)
Energy = Power* time
Energy = I*V*t
Energy = 135 * 12 * 3600 s
Energy = 5.8 * 10^6 J
(c)
mass of battery , M = 18.1 Kg
energy stored = V * I * t
m * c^2 = 12 * 135 * 3600
m = 6.48 *10^-11 kg
fraction of mass lost = m/M
fraction of mass lost = ( 6.48 *10^-11 )/18.1
fraction of mass lost = 3.58 *10^-12
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(d)
let the number of batteries is N
N * 12 * 135 * 3600 = 14 * 1.32 *10^8 * 2.5
solving for N
N = 792
mass of battery neeed = 792 * 18.1 Kg
mass of battery neeed = 14335.2 Kg
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