Thermo and Projectile Motion: Bathtub Bather Launch Calculation

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Homework Statement


A person of mass 70.0 kg is sitting in the bathtub. The bathtub is 190.0 cm by 80.0 cm before the person got in, the water was 10.0 cm deep. The water is at a temperature of 37.0 C. Suppose that the water were to cool down spontaneously to form ice at 0.0 C, and that all the energy released was used to launch the hapless bather vertically into the air.

Homework Equations



The Attempt at a Solution



Density of Water = 1000 kg/m^3

Volume of water
1.9 * 0.8 * 0.1 = 0.152 m^3

Mass of water
0.152 m^3 * 1000 kg/m^3 = 152 kg

Energy of cooling water = mcT
152 * 4190 * (0-37) = 2.356456 x 10^7

Energy of freezing water = mL
-152 * 3.34*10^5 =-5.092 x 10^7

Sum of Energy
-7.433256 x 10^7nvm got it..
h=KE/mg
 
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