Thermo and Projectile Motion: Bathtub Bather Launch Calculation

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SUMMARY

The discussion focuses on calculating the energy transfer involved in launching a person from a bathtub as water cools and freezes. A person weighing 70.0 kg sits in a bathtub with a water volume of 0.152 m³ at 37.0°C. The energy released from cooling the water to 0.0°C is calculated to be approximately 2.36 x 10^7 J, while the energy from freezing the water is about -5.09 x 10^7 J. The total energy available for launching the bather is approximately -7.43 x 10^7 J, leading to a discussion about gravitational potential energy and its relation to kinetic energy.

PREREQUISITES
  • Understanding of basic thermodynamics, specifically energy transfer and phase changes.
  • Familiarity with gravitational potential energy and kinetic energy equations.
  • Knowledge of water density and specific heat capacity.
  • Basic algebra for manipulating equations and solving for variables.
NEXT STEPS
  • Study the principles of thermodynamics, focusing on energy conservation and phase transitions.
  • Learn about gravitational potential energy and its calculation in physics.
  • Explore the specific heat capacity of various substances, particularly water.
  • Investigate real-world applications of energy transfer in fluid dynamics.
USEFUL FOR

Students studying physics, particularly those focusing on thermodynamics and mechanics, as well as educators looking for practical examples of energy calculations in real-life scenarios.

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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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