Calorimetry: Thermal energy going into translational KE

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SUMMARY

The discussion centers on calculating the speed at which a cup of boiling water (250g) would fly off a table after cooling to room temperature (25°C), with the released thermal energy converted into translational kinetic energy (KE). The participant calculated the thermal energy (Q) using the formula Q=mcΔT, resulting in -78487.5 J after conversion from calories. Adjusting for degrees of freedom, the effective energy was halved to 39243.75 J, leading to a final velocity of 560 m/s using the kinetic energy formula KE = (1/2)mv².

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  • Understanding of thermodynamics, specifically the first law of thermodynamics.
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  • Knowledge of unit conversions between calories and joules.
  • Basic grasp of degrees of freedom in molecular motion.
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  • Explore advanced kinetic theory of gases and its applications.
  • Investigate the relationship between thermal energy and kinetic energy in different states of matter.
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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
Uwater=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?
 
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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?
 
Any help?
 

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