The Amount of Energy released by a tropical thunderstorm

AI Thread Summary
The discussion focuses on the relationship between precipitation and energy in tropical thunderstorms, emphasizing the latent heat released during condensation. It highlights that the mass of water in a storm is 9.72 * 10^8 kg and involves calculations based on the storm's air intake and condensed water volume. The latent heat of vaporization is specified as 2.5 * 10^6 J/kg, which is crucial for determining the total energy output. Participants are encouraged to calculate the total Joules of energy generated by the storm using these parameters. Understanding this energy release is essential for comprehending thunderstorm dynamics.
button2
Messages
4
Reaction score
0
I can't quite figure this out... but given this sceneraio:
Precipitation and energy can be thought of as closely related quaniities in a thunderstorm. That is, when rain forms via condensation, the atmosphere warms from the latent heat released. The amount of heat energy must be proportional to the amount of rainfall produced. In this case, the mass of the water content in a given thunderstorm is 9.72 * 10^8 kg. If a storm ingests 2.16*10^9m^3 of air with 9.72*10^5m^3 as condensed water, and given that the latent heat of vaporization is 2.5*10^6J/kg, what is the total number of Joules of energy generated by this storm? :confused:
 
Physics news on Phys.org
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top