The battery of a mobile phone is rated as 4.2 V, 5800 mAh. How much energy is stored in it when fully charged?

The battery is specified by two numbers: voltage $$V = 4.2\text{ V}$$ and capacity $$C = 5800\text{ mAh}$$.

Step 1 Convert the capacity into ampere-hours (Ah)
Since $$1000\text{ mAh} = 1\text{ Ah}$$, we have
$$C = \frac{5800}{1000}\text{ Ah} = 5.8\text{ Ah}$$

Step 2 Convert ampere-hours to coulombs (C)
By definition, $$1\text{ A} = 1\text{ C s}^{-1}$$ and there are $$3600\text{ s}$$ in one hour. Hence
$$1\text{ Ah} = 1\text{ A}\times3600\text{ s} = 3600\text{ C}$$
Therefore,
$$Q = 5.8\text{ Ah}\times3600\text{ C Ah}^{-1} = 20880\text{ C}$$

Step 3 Calculate the stored energy
Electrical energy stored, $$E = VQ$$, so
$$E = 4.2\text{ V}\times20880\text{ C} = 87696\text{ J}$$

Step 4 Express the result in kilojoules
$$1\text{ kJ} = 10^{3}\text{ J}$$, hence
$$E = \frac{87696}{10^{3}}\text{ kJ} \approx 87.7\text{ kJ}$$

Thus, the energy stored in the fully charged mobile phone battery is about $$87.7\text{ kJ}$$.

Option C is correct.