An iron meteorite with a mass of 4,000 kg and a temperature of 1,300ºC is travel

Question

An iron meteorite with a mass of 4,000 kg and a temperature of 1,300ºC is traveling at a speed of 4.5 km/s when it crash lands into a pool of water on Earth. There are 30.0 m3 of water in the pool, and the water temperature is 20.0ºC. Let’s assume that half of the water in the pool is splashed out and doesn’t have time to absorb any heat. The remaining water stays in the pool and does absorb energy and heat from the meteorite. Also assume that 15% of the meteorite’s kinetic energy is absorbed by the water that stays in the pool and the remaining 85% of the meteorite’s kinetic energy is transferred into the ground (and the water that was splashed out of the pool). After coming to rest, the meteorite is in thermal contact with the water in the pool and there is thermal transfer of energy between the meteorite and the water due to their temperature difference. Once the meteorite and the remaining water in the pool have reached equilibrium, what is the final temperature of the water and meteorite, and how much water if any has evaporated? (Ignore any heat absorbed by the walls of the pool and by the air.)

Explanation / Answer

remaining water in pool is mw = (1/2)*density*volume = (1/2)*1000*30 = 15000 kg

Total energy given to the remanining water in pool is Q

Q = 0.15*m*v^2 + (mW*sW*dT) = m*si*dT

sw is the specific heat capacity of water = 4186 J/kg-K

si is the specific heat capacity of iron = 450 J/kg-K

So 0.15*m*v^2 + (mW*sW*dT) = m*si*dT

(0.15*4000*4500^2)+(15000*4186*(T-20)) = 4000*450*(1300-T)

10894200000+(62790000*T) = 2340000000-1800000T

T = -132.43 degrees celsius

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