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Q40 PE

Expert-verifiedFound in: Page 262

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**(a) What is the available energy content, in joules, of a battery that operates a 2.00-W electric clock for 18 months? **

**(b) How long can a battery that can supply 8.00×10 ^{4} J run a pocket calculator that consumes energy at the rate of 1.00×10^{−3} W?**

(a) The energy content of available for the battery is \(93.31 \times {10^6}{\rm{ J}}\).

(b) The battery can be used up to \(2.54{\rm{ years}}\).

**Power: Power is a scalar quantity which is defined how fast the energy is being used. **

Mathematically,

\(P = \frac{E}{T}\) (1.1)

Here, E stands for energy, and T stands for time.

(a)

The expression for the energy content of the battery can be calculated using equation (1.1).

Rearranging equation (1.1) in order to get an expression for the energy content.

\(E = PT\)

Here, P is the power of battery \(\left( {P = 2.00{\rm{ W}}} \right)\) and T is the time \(\left( {T = 18{\rm{ months}}} \right)\).

Putting all known values,

\(\begin{aligned}E &= \left( {2.00{\rm{ W}}} \right) \times \left( {18{\rm{ month}}} \right)\\ &= \left( {2.00{\rm{ W}}} \right) \times \left( {18{\rm{ month}}} \right) \times \left( {\frac{{30{\rm{ day}}}}{{1{\rm{ month}}}}} \right) \times \left( {\frac{{24{\rm{ hr}}}}{{1{\rm{ day}}}}} \right) \times \left( {\frac{{3600{\rm{ s}}}}{{1{\rm{ h}}}}} \right)\\ &= 93.31 \times {10^6}{\rm{ J}}\end{aligned}\)

Therefore, the energy content of available for the battery is \(93.31 \times {10^6}{\rm{ J}}\).

(b)

The life of the battery is,

\(T = \frac{E}{P}\)

Here, *E* is the energy contain of the battery \(\left( {8.00 \times {{10}^4}{\rm{ J}}} \right)\), and P is the power \(\left( {1.00 \times {{10}^{ - 3}}{\rm{ W}}} \right)\).

Putting all known values,

\(\begin{aligned}T &= \frac{{8.00 \times {{10}^4}{\rm{ J}}}}{{1.00 \times {{10}^{ - 3}}{\rm{ W}}}}\\ &= \left( {8.00 \times {{10}^7}{\rm{ sec}}} \right) \times \left( {\frac{{1{\rm{ hr}}}}{{3600{\rm{ s}}}}} \right) \times \left( {\frac{{1{\rm{ day}}}}{{24{\rm{ hr}}}}} \right) \times \left( {\frac{{1{\rm{ year}}}}{{365\,{\rm{day}}}}} \right)\\ &= 2.54{\rm{ years}}\end{aligned}\)

Therefore, the battery can be used up to \(2.54{\rm{ years}}\).

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