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Q40 PE

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Found in: Page 262

### College Physics (Urone)

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 8.00×104 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 $$2.54{\rm{ years}}$$.

See the step by step solution

## Definition of Concepts

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.

## Find the energy content of the battery

(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 $$93.31 \times {10^6}{\rm{ J}}$$.

## Find the life of the battery

(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 $$2.54{\rm{ years}}$$.