A .5 kg rock is dropped from a height of 20 meters into a pail

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SUMMARY

The discussion revolves around a physics problem involving a 0.5 kg rock dropped from a height of 20 meters into a pail containing 0.6 kg of water. The key equation to solve the problem is based on the principle of energy conservation, specifically the conversion of potential energy (PE) to heat energy (Q) in both the rock and water. The correct formulation is mgh + mcT(rock) = mcT(water), where m is mass, g is acceleration due to gravity, h is height, c is specific heat, and T is temperature change. The discussion emphasizes the importance of understanding energy transfer between the rock and water, particularly when they start at different temperatures.

PREREQUISITES
  • Understanding of potential energy (PE) and kinetic energy (KE)
  • Knowledge of specific heat capacity and its calculation
  • Familiarity with the conservation of energy principle
  • Basic algebra for solving equations
NEXT STEPS
  • Study the concept of energy conservation in thermodynamics
  • Learn how to calculate specific heat and temperature change using Q = mcΔT
  • Explore examples of energy transfer in different states of matter
  • Investigate the effects of varying initial temperatures on heat exchange
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding energy transfer principles in thermodynamics and mechanics.

Sheldinoh
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My Physics teacher gave us a quiz problem and I don't understand it really how he got his answer. Can you please give me your answer and an explanation for the answer Thanks. Here is the question:

A .5 kg rock is dropped from a height of 20 meters into a pail containing .6 kg of water. The rock has a specific heat of 1480 and the water has a specific heat of 4186. What is the is the rise in temperature of the rock and water ?
 
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Show what you have tried already.
 


I think it is:
mgh(energy of fall ) = mcT + mcT


My teacher says its:
mgh + mcT = mcT
 


.In terms of energy changes PE changes to KE and then to heat in both the water and the rock.It is not necessary to calculate the KE so go straight to your equation which is the correct one.Make sure you plug in the right values for m and c.
 


Thanks for agreeing with me. Could you help me out now with making my physics teacher agree with me. He says that you add the energy of the rock with the heat from the rock to equal the heat of the water. mgh+mcT(rock)=mcT(water). I think it is wrong but I have no clue how to explain it. Do you know how.
 


The impression gained from your question was that the rock and water were both at the same temperature at the start. Let us assume they are not.
If the rock were colder then it would gain some of its heat energy from the heat energy of the water and if the water were colder it would gain some of its heat energy from the heat energy rock.If they were at the same temperature, which is what I assume from your question,then both would gain heat energy from the PE only and both of them would heat up.
 

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