A rock of mass 0.19 kg falls from rest from a height of 15 m into a pail containing

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Thermodynamics

A rock of mass 0.19 kg falls from rest from a height of 15 m into a pail containing 0.37 kg of water. The rock and water have the same initial temperature. The specific heat capacity of the rock is 1885 J/(kg·C°). Ignore the heat absorbed by the pail itself, and determine the rise in the temperature of the rock and water.

I listed by knowns and unknowns. The rock is of mass .19kg and the distance is 15m. The mass of water is .37kg. The Tinitial is the same for both water ad rock. Heat capacity of the rock is 1885.

I have no idea how to solve this except that it involves using the formula Q=cm(delta T)

Please give me a solution. Thanks
 
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ehild
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Assume that the mechanical energy of the rock is converted to heat.

ehild
 
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so does this mean to use a mechanical energy equation.
 
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ehild
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Yes, assume that the mechanical energy of the rock is conserved during the fall. When it splashes into the water, the water is brought into chaotic motion and the energy of this motion is transformed into the energy of the random motion of molecules, that is, heat.
 
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PE=mgh
KE= .5(m)(v)^2
E=PE+KE (I am not sure if this is the right equation for mechanical energy?)
 
  • #6
ehild
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PE=mgh
KE= .5(m)(v)^2
E=PE+KE (I am not sure if this is the right equation for mechanical energy?)
You can use E=PE+KE for a certain position of the rock. If it is at rest at height of 15 m, does it have kinetic energy?
When it fall down to the ground what is the potential energy?

ehild
 
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no kinetic energy
 
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ehild
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So what is the mechanical energy of the rock?

ehild
 
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just potential energy
 
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ehild
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now my head really hurts. i have no idea. can i get a detalied step by step solution. this thing is due by 6:00
 
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A rock of mass 0.19 kg falls from rest from a height of 15 m into a pail containing 0.37 kg of water. The rock and water have the same initial temperature. The specific heat capacity of the rock is 1885 J/(kg·C°). Ignore the heat absorbed by the pail itself, and determine the rise in the temperature of the rock and water.

I listed by knowns and unknowns. The rock is of mass .19kg and the distance is 15m. The mass of water is .37kg. The Tinitial is the same for both water ad rock. Heat capacity of the rock is 1885.

I have no idea how to solve this except that it involves using the formula Q=cm(delta T)

Please give me a solution. Thanks
Since the height of fall is small, we can approximate with PE = mgh

Considering that all PE is converted to KE which subsequently converts to the total increase in heat energy,

PE = Heat(Rock) + Heat (Water
(0.19)(15)(9.81) = (0.19)(1885)(change temp) + 0.37(4200)(change temp)

By switching things around, you should get the answer.
 

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