Question

A fridge uses about 200 watts of power. This means that the fridge uses 200 joules per second of energy. If managed to turn 1 g of matter into energy, according to Einstein's equation E = mc², how long would I be able to power my fridge with that energy?​

Answer 1

Approximately
`4.50 * 10^(11)\; {\rm s}`.

Step-by-step explanation:

The speed of light is
`c \approx 3.00* 10^(8)\; {\rm m\cdot s^(-1)}`.

Note that the standard unit of energy, joule, is a derived unit. In terms of the standard base units:

`\begin{aligned}1\; {\rm J} &= (1\; {\rm N})\, (1\; {\rm m}) \\ &= (1\; {\rm kg \cdot m \cdot s^(-2)})\, (1\; {\rm m}) \\ &= 1\; {\rm kg \cdot m^(2) \cdot s^(-2)} \end{aligned}`.

Apply unit conversion and ensure that the unit of mass is in the standard unit kilogram (
`{\rm kg}`):

`\begin{aligned} m &= 1\; {\rm g} * \frac{1\; {\rm kg}}{10^(3)\; {\rm g}} &= 10^(-3)\; {\rm kg}\end{aligned}`.

Apply the equation
`E = m\, c^(2)` to find the energy equivalent to
`m = 10^(-3)\; {\rm kg}` of matter:

`\begin{aligned}E &= m\, c^(2) \\ &= (10^(-3)\; {\rm kg})\, (3.00 * 10^(8)\; {\rm m\cdot s^(-1)})^(2) \\ &= 9.00* 10^(13)\; {\rm kg \cdot m^(2) \cdot s^(-2)} \\ &= 9.00* 10^(13)\; {\rm J} \end{aligned}`.

Divide energy
`E` by power
`P` to find the duration
`t` of the power consumption:

`\begin{aligned} t &= (E)/(P) \\ &\approx \frac{9.00 * 10^(13)\; {\rm J}}{200\; {\rm J \cdot s^(-1)}} \\ &= 4.50 * 10^(11)\; {\rm s}\end{aligned}`.

In other words, the energy equivalent to
`1\; {\rm g}` of matter could power this fridge for approximately
`4.50 * 10^(11)\; {\rm s}`.