Calorimetry: Thermal energy going into translational KE

[lolcat]

Homework Statement

Suppose a cup of boiling water (m=250g) instantaneously cools to room temperature (25°C) with the liberated thermal energy going into translational KE. How fast will the cup fly off the table? Assume the water molecules have 18 degrees of freedom.

Homework Equations

Q=mcΔT
U_water=9nRT (not sure about this one)

The Attempt at a Solution

So what I first did was try to calculate Q:

Q=mcΔT
Q=(250g)(1)(25-100)
Q=-18750 cal

I then converted calories to joules:

-18750 cal x 4.186 J = -78487.5 J.

I am not sure where to go after this. Any help?

 

[lolcat]

Update:

I'm guessing that since the cup is on a horizontal table, the degrees of freedom is 9. Therefore the energy must be halved.

Q = 39243.75 J

Plug it into:

KE = (1/2)mv²

v² = (2KE) / m
= (2*39243.75) / .250
= 313950

√v² = √313950
v = 560 m/s

Can anyone confirm that this is correct?

 

[lolcat]

Any help?