Answer:
Approximately [tex]4.50 \times 10^{11}\; {\rm s}[/tex].
Explanation:
The speed of light is [tex]c \approx 3.00\times 10^{8}\; {\rm m\cdot s^{-1}}[/tex].
Note that the standard unit of energy, joule, is a derived unit. In terms of the standard base units:
[tex]\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}[/tex].
Apply unit conversion and ensure that the unit of mass is in the standard unit kilogram ([tex]{\rm kg}[/tex]):
[tex]\begin{aligned} m &= 1\; {\rm g} \times \frac{1\; {\rm kg}}{10^{3}\; {\rm g}} &= 10^{-3}\; {\rm kg}\end{aligned}[/tex].
Apply the equation [tex]E = m\, c^{2}[/tex] to find the energy equivalent to [tex]m = 10^{-3}\; {\rm kg}[/tex] of matter:
[tex]\begin{aligned}E &= m\, c^{2} \\ &= (10^{-3}\; {\rm kg})\, (3.00 \times 10^{8}\; {\rm m\cdot s^{-1}})^{2} \\ &= 9.00\times 10^{13}\; {\rm kg \cdot m^{2} \cdot s^{-2}} \\ &= 9.00\times 10^{13}\; {\rm J} \end{aligned}[/tex].
Divide energy [tex]E[/tex] by power [tex]P[/tex] to find the duration [tex]t[/tex] of the power consumption:
[tex]\begin{aligned} t &= \frac{E}{P} \\ &\approx \frac{9.00 \times 10^{13}\; {\rm J}}{200\; {\rm J \cdot s^{-1}}} \\ &= 4.50 \times 10^{11}\; {\rm s}\end{aligned}[/tex].
In other words, the energy equivalent to [tex]1\; {\rm g}[/tex] of matter could power this fridge for approximately [tex]4.50 \times 10^{11}\; {\rm s}[/tex].
