Different forms of energy

Radioactive decay of radium-226 : We know, that the luminous figures have an decay rate of $\dot N(0) = -40,000 \,\text{s}^{-1}$ . If we use the law for radioactive decay, we can calculated the total amount of radium atoms: $\begin{aligned} N(t) &= N_0 \exp\left(- \frac{\ln(2)}{t_{1/2}} \cdot t \right) \\ \Rightarrow \quad \dot N(t) &= - \frac{\ln(2)}{t_{1/2}} N_0 \exp\left(- \frac{\ln(2)}{t_{1/2}} \cdot t \right) \\ \Rightarrow \quad N_0 &= -\frac{t_{1/2}}{\ln(2)} \dot N(0) = \frac{1600 \cdot 365 \cdot 24 \cdot 60 \cdot 60}{\ln(2)} \cdot 40,000 \approx 2.9 \cdot 10^{15} \,\text{atoms} \end{aligned}$ with the half-life $t_{1/2} = 1600\,\text{years}$ . Each atom releases on decay an energy of $E = 58 \,\text{MeV} = e \cdot 58 \,\text{MV}$ with the elemental charge $e \approx 1.6 \cdot 10^{-19} \,\text{C}$ . The total energy results $E_\text{nuclear} = N_0 E \approx (2.9 \cdot 10^{15}) \cdot (58 \cdot 10^{6}) \cdot (1.6 \cdot 10^{-19}) \,\text{kJ} \approx 27 \,\text{kJ}$ Note: Even if such a watch releases radiation for a long period of time and most of the radiation does not hit the body, you should not wear it on your wrist all the time. In modern watches, however, either the much less dangerous tritium is used or radioactive elements are completely eliminated.

AA batteries : The batteries have a total charge of $Q = 3 \,\text{Ah} = 10.8 \,\text{kC}$ and a voltage $U = 1.5 \,\text{V}$ , so that the electric energy results $E_\text{el} = 4 \cdot Q \cdot U \approx 65 \,\text{kJ}$ (We assumed, that the voltage remains constant during the discharge in contrast to a capacitor.)

Baseball : The sound velocity in air is approximatly $c \approx 340\,\text{m}/\text{s}$ , so that the kinetic energy results $E_\text{kin} = \frac{1}{2} m (4c)^2 \approx 139 \,\text{kJ}$

Person on top of the Empire State Building : With an average weight $m \approx 80\,\text{kg}$ of an adult we get an potential energy $E_\text{pot} = m g h \approx 80 \cdot 9.8 \cdot 443 \,\text{J} \approx 350 \,\text{kJ}$

Pot of coffee : If we assume a temperature of $T_\text{hot} \approx 90^\circ \text{C}$ for the coffee water and a room temperature of $T_\text{room} \approx 20 ^\circ \text{C}$ , so that $\Delta T \approx 70^\circ \text{C}$ Since the heat capacity of one liter water equals $C = 1 \text{kcal}/^\circ \text{C}$ we get an total thermal energy of $E_\text{therm} = m \cdot C \cdot \Delta T \approx 1.5 \cdot 4187 \cdot 70 \,\text{J} \approx 440\,\text{kJ}$

Chocolate : Chocolate has an calorific value of $H \approx 500 \,\text{kcal}/\,100 \text{g}$ . Half a chocolate bar has therefore the chemical energy of $E_\text{chem} = m \cdot H = \frac{1}{2} \cdot 500 \cdot 4187\,\text{J} \approx 1050 \,\text{kJ}$

and the final winner is..

The scuba tank: If caculate the inner energy of the compressed air in the scuba tank, we get $U = \frac{5}{2} n R T = \frac{5}{2} p V = \frac{5}{2} \cdot (300 \cdot 10^{5}) \cdot (12 \cdot 10^{-3}) \,\text{J} = 900 \,\text{kJ}$ which would be less than the energy of the chocolate. But we asked for the thermodynamic work, that can be released, if the air expands to normal pressure. This energy value is higher, because the compressed air absorbs heat from the environment during the expansion. For isothermal expansion we get $E = \int_{V}^{V'} p dV = \int_{V}^{V'} \frac{n R T}{V} dV = n R T \log \frac{V'}{V} = p V \log \frac{p}{1 \,\text{bar}} = (300 \cdot 10^{5}) \cdot (12 \cdot 10^{-3}) \cdot log(300) \approx 2050 \,\text{kJ}$ since $pV = \text{const}$ during expansion.