Spaceship holed in one?

Question

A bit of space debris penetrates the hull of a spaceship traversing the asteroid belt and comes to rest in a container of water that was at 0oC before being hit. The mass of the space rock is 0.92g and the mass of the water is 1.15kg. If the space rock traveled at 8.17 x 103 m/s, and all of its kinetic energy was used to heat the water, what is the final temprature of the water?
Assume the interaction happens fast enough that the water does not boil.

Answer 1

Gee... I did not know water had such 'holding' power. :-)
Assuming that all kinetic energy Ke of the space junk turns into heat Q we have
Q=Ke
Ke= 0.5m1V^2
Energy to make the water boil
Q1=m2Cp(T2-T1)
Cp = 4.1813 J/(g K)
Q1=1150 4.1813 p(100-0)=
Q1=481,000 J

Ke= 0.5m1V^2
Ke= 0.5 (.92 E-3) x(8.17 E+3)^2
Ke= 3,070,000 Joules (May I say WOW! Thank you)

The water would turn to steam

total energy
Qt=Q1+Q2+Q3
Q1- to heat water from 0 - 100 deg C
Q2- To evaporate water
Q3- heat the steam
Q2= mCv=1150 x 2272=2,610,000
Q3= mCv(T2-T1) T1= 100 deg C or 372 deg K

Energy left to heat the steam Q'
Q'=Ke - [Q1+ Q2]=
Q'= 3,070,000 -[481,000 + 2,610,000]
Q'= -21000

Looks like almost all that water turns to steam leaving
m'=q'/Cv= 21000/2272 = 9.2gm of water in liquid form but at 100 degrees C.

The physics is a bit more complex since steam would try to expand and pressure inside the bottle would be sufficient for the bottle to burst with boiling hot water. The strength of the container will be the function of amount of water converted.