### Far Horizon

An observer is on top of a lighthouse. How far from the foot of the lighthouse is the horizon that the observer can see?

### Investigating the Dilution Series

Which dilutions can you make using only 10ml pipettes?

### chemNRICH

chemNRICH is the area of the stemNRICH site devoted to the mathematics underlying the study of chemistry, designed to help develop the mathematics required to get the most from your study of chemistry at A-level and university.

# Global Warming

##### Age 14 to 16 Challenge Level:

According to wikipedia, the atmosphere has a mass of $5\times10^{18}kg$, and the specific heat capacity of air is about $1\mathrm{Jg^{-1}K^{-1}}$. Therefore, the amount of energy needed to raise the average temperature of the atmosphere by $0.4^{\circ}\mathrm{C}$ is
$$1\times10^3\mathrm{Jkg^{-1}K^{-1}}\times0.4\mathrm{K}\times5\times10^{18}\mathrm{kg} = 2\times10^{21}\mathrm{J}.$$

Coal has an energy density of 24 megajoules per kilogram. Assuming 100% of the energy released from burning heats the atmosphere, we'll need $\frac{2\times10^{21}\mathrm{J}}{ 24\times10^6\mathrm{Jkg^{-1}}} = 8\times10^{13}\mathrm{kg}$. Assuming a global population of 6 billion this corresponds to $\frac{8\times10^{13}}{6\times10^9\times30\times52} = 9$ kilograms per person per week per year for the last 30 years.

This is probably a significant underestimate of the amount of fuel used, as not all the energy released from burning goes straight to the atmosphere. Also, this calculation ignores the fact that different fuels have different energy densities.

The greenhouse effect is also responsible for some of the temperature increase.