Question

Alkaline batteries have the advantage of putting out constant voltage until very nearly the end of their life. How long will an alkaline battery rated at \(1.00\;A/h\) and \(1.58\;V\)keep a \(1.00 - W\) flashlight bulb burning?

Short answer

It takes \(1.58\;h\) for an alkaline battery rated at \(1.00\;A/h\) and \(1.58\;V\) to keep a\(1.00\;W\) flashlight bulb burning.

Step-by-step solution

Definition of Power

The power of a process is the amount of some type of energy converted into a different type divided by the time interval\(\Delta t\)in which the process occurred:

\(P = \frac{{\Delta E}}{{\Delta t}}......(1)\)

The SI unit of power is the watt\((W)\).

\(1\) Watt is\(1\)joule/second\((1\;W = 1\;J/s)\).

Definition of Potential energy

The potential energy\(\Delta U\)expression is given by

\(\Delta U = q\Delta V......(2)\)

Where\(q\)is the change, and\(\Delta V\)is the potential difference.

The given data

• The charge moved by the battery is: \(q = (1.00\;A/h)\left( {\frac{{3600\;s}}{{1\;h}}} \right) = 3600{\rm{ }}C\).
• The potential difference across the battery is: \(\Delta V = 1.58\;V\).
• The power consumed by the flashlight is: \(P = 1.00\;W\).

Calculation of the energy consumed

The energy supplied by the battery, from Equation \((2)\):

\(E = q\Delta V\)

Substituting the values for \(q\)and \(\Delta V\)

\(\begin{align}{}E &= (3600C)(1.58\;V)\\ &= 5688\;J\end{align}\)

The supplied energy is \(5688\;J\).

Calculation of the power consumed

The power consumed by the flashlight, from Equation \((1)\):

\(P = \frac{E}{{\Delta t}}\)

Solve for \(\Delta t\)as,

\(\Delta t = \frac{E}{P}\)

Substitute the values:

\(\begin{align}{}\Delta t &= \frac{{5688\;J}}{{1.00\;W}}\\ &= 5688\;s\\ &= (5688\;s)\left( {\frac{{1\;\min }}{{60\;s}}} \right)\left( {\frac{{1\;h}}{{60\;\min }}} \right)\\ &= 1.58\;h\end{align}\)

Therefore, the value for time is obtained as \(1.58\;h\).