Energy Conservation Dilemma Assume that a man is travelling in a space ship at a certain relativistic speed with respect to a man at rest at some point in space, such that 3 minutes in the ship is equal to 5 minutes for the person at rest.
Also assume that the man in the ship has a lighter which contains gas of a certain amount such that the lighter can be lit for 5 minutes .
Now if the man in the space ship lights the lighter for 3 minutes, then he would have 2 minutes' worth of gas left, but the stationary observer would have seen light emitted for about 5 minutes (since 3 in that space ship = 5 minutes for the stationary observer)
How is it possible for the stationary observer to see light for 5 minutes?
And in this case, how is energy conserved?
 A: Following on from @Jahan Claes' answer we can do some maths!
Say the man on the spaceship is using an ethanol burner. Ethanol has an enthalpy change of $1058 \text{ kJ}/\text{mol}$ but for simplicity let's say that it's equal to $1050$. If he burns $1$ mole of ethanol then it gives out approximately $1050 \text{ kJ}$. For the moving observer this is equal to $350\,000$ joules/min. For the stationary observer this will be equal to $210\,000$ joules/min.
Let's also say that the ethanol burner is giving out orange light, of a wavelength of $600 \text{ nm}$. Using the equation $$E = \frac{hc}{\lambda}$$ we can work out the energy of each photon to be roughly $3.315\times 10^{-19}$ (luckily for us the speed of light is the same for all observers). This means that the moving observer sees approximately $1.056\times 10^{24}$ photons every minute, whilst the stationary observer sees only approximately $6.335\times 10^{23}$ photons every minute.
From this we can see that $1.056(...)\times 10^{24}\times3 =  6.335(...)\times 10^{23}\times 5$ and everything works out just fine :)
A: From the perspective of the stationary observer, the light is dimmer. Why? Because the chemical reaction is happening more slowly, so the fire emits fewer photons per second. The flame burns for longer, but it emits less energy per second. Both observers will agree on the total energy emitted by the flame (once they have accounted for possible red-shift) and thus total energy is conserved.
A: You say the gas in the lighter is sufficient to be lit for 5 minutes, but whose 5 minutes? I guess you mean 5 mins as reckoned by the spaceship observer. In that case, for the observer at rest, the amount of gas in lighter is sufficient to be lit for $5\times (5/3)=8.3$ mins. So when spaceship observer lights the gas only for 3 mins and has 2 mins worth of gas left over, observer at rest sees that the gas has been lit for 5 mins and 3.3 mins worth of gas is left over. There is no contradiction or violation of any law.
P.S. If you say that the amount of gas in the lighter can burn for 5 mins as per observer at rest, then from spaceship observer's point of view the lighter contains only 3 mins worth of gas. Then if he lights it for 3 mins there cannot be any gas left inside the lighter.
