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Quantity of heat

  1. Oct 15, 2006 #1
    This is one of those test your understanding questions out of the book.

    What distance would a 1.00kg block have to fall to raise the temperature of .250kg of water from 20C to 90C? Assume that all potential emergy that the block loses as it falls goes into raising the emperature of the water?
    (The system: the block is suspended by string through a pully and t attached to a axel. The axel has more thread wrapped around it and is attached to a horizontally rotating paddle inside a glass filled with water. so as the block falls the paddle is doing work on the water resulting in heat transfer Q. this is the most accurate description I can come up with from looking at the picture)

    This is the answer right out of the book:

    The temperature change is T = 70K so the amount of heat is Q = (m_water)(c)(dT) = (1kg)(4190J/kgK)(70K) = 2.93x10^5J If the block falls a distance h, the amount of potential energy lost is m_block(g)(h), so h = Q/(m_block)(h) = (2.93e+10^5J)/(1.00kg)(9.8m/s^2) = 2.99^4 m

    Is this a typo? They clearly distinguish m_water and m_block, yet they use 1kg for the mass of the water instead of .250kg. Or am I misunderstanding the question? I get Q = (.250)(c)(dt).
  2. jcsd
  3. Oct 15, 2006 #2


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    It's clearly a typo. You are right, the mass of the water should be 0.25 kg in the expression.
  4. Oct 15, 2006 #3
    thanks alot!!
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