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Stored energy and potential energy

What happens to energy stored in an object at rest, such as the energy put into a battery by charging it, or the energy stored in a toy gun’s compressed spring? The energy input becomes part of the total energy of the object and thus increases its rest mass. All stored and potential energy becomes mass in a system. In seeming contradiction, the principle of conservation of mass (meaning total mass is constant) was one of the great laws verified by nineteenth-century science. Why was it not noticed to be incorrect? The following example helps answer this question.

Calculating rest mass

A car battery is rated to be able to move 600 ampere-hours ( A · h ) of charge at 12.0 V. (a) Calculate the increase in rest mass of such a battery when it is taken from being fully depleted to being fully charged, assuming none of the chemical reactants enter or leave the battery. (b) What percent increase is this, given that the battery’s mass is 20.0 kg?

Strategy

In part (a), we first must find the energy stored as chemical energy E batt in the battery, which equals the electrical energy the battery can provide. Because E batt = q V , we have to calculate the charge q in 600 A · h, which is the product of the current I and the time t . We then multiply the result by 12.0 V. We can then calculate the battery’s increase in mass using E batt = ( Δ m ) c 2 . Part (b) is a simple ratio converted into a percentage.

Solution for (a)

  1. Identify the knowns: I · t = 600 A · h; V = 12.0 V; c = 3.00 × 10 8 m/s.
  2. Identify the unknown: Δ m .
  3. Express the answer as an equation:
    E batt = ( Δ m ) c 2 Δ m = E batt c 2 = q V c 2 = ( I t ) V c 2 .
  4. Do the calculation:
    Δ m = ( 600 A · h ) ( 12.0 V ) ( 3.00 × 10 8 ) 2 .

    Write amperes A as coulombs per second (C/s), and convert hours into seconds:
    Δ m = ( 600 C/s · h ) ( 3600 s 1 h ) ( 12.0 J/C) ( 3.00 × 10 8 m/s) 2 = 2.88 × 10 −10 kg.

    where we have used the conversion 1 kg · m 2 /s 2 = 1 J .

Solution for (b)

For part (b):

  1. Identify the knowns: Δ m = 2.88 × 10 −10 kg; m = 20.0 kg.
  2. Identify the unknown: % change.
  3. Express the answer as an equation: % increase = Δ m m × 100 % .
  4. Do the calculation:
    % increase = Δ m m × 100 % = 2.88 × 10 −10 kg 20.0 kg × 100 % = 1.44 × 10 −9 % .

Significance

Both the actual increase in mass and the percent increase are very small, because energy is divided by c 2 , a very large number. We would have to be able to measure the mass of the battery to a precision of a billionth of a percent, or 1 part in 10 11 , to notice this increase. It is no wonder that the mass variation is not readily observed. In fact, this change in mass is so small that we may question how anyone could verify that it is real. The answer is found in nuclear processes in which the percentage of mass destroyed is large enough to be measured accurately. The mass of the fuel of a nuclear reactor, for example, is measurably smaller when its energy has been used. In that case, stored energy has been released (converted mostly into thermal energy to power electric generators) and the rest mass has decreased. A decrease in mass also occurs from using the energy stored in a battery, except that the stored energy is much greater in nuclear processes, making the change in mass measurable in practice as well as in theory.

Practice Key Terms 4

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Source:  OpenStax, University physics volume 3. OpenStax CNX. Nov 04, 2016 Download for free at http://cnx.org/content/col12067/1.4
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