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Find the temperature change of a person struck by lightning

  1. Dec 7, 2014 #1
    1. The problem statement, all variables and given/known data
    A lightning flash releases about 1010J of electrical energy.

    If all this energy is added to 50 kg of water (the amount of water in a 165-lb person) at 37∘C, what are the final state and temperature of the water? The specific heat of water is 4180 J/kg⋅∘C, heat of vaporization at the boiling temperature for water is 2.256×106J/kg, the specific heat of steam is 1970 J/kg⋅∘C

    Find ΔT in ∘C

    2. Relevant equations
    ΔU = Q - W
    Q = mCΔT
    Q = mL

    3. The attempt at a solution
    I found
    Qwater = (50kg)(4180J/kg⋅∘C)(100-37) = 1.13167⋅10^7 J
    Qphase change to steam = (50kg)(2.256×106J/kg) = 1.128⋅10^8 J
    Qtotal = 1.13167⋅10^7 J + 1.128⋅10^8 J = 1.25967⋅10^8 J

    I'm not sure what to do with the energy from the lightning strike in order to find the total change in temperature.
     
  2. jcsd
  3. Dec 7, 2014 #2
    I can't figure out how to delete this thread, but I found out I didn't need any of that information or calculations. using Q=mCΔT > ΔT=Qlightning/Cm = 1*10^5 ∘C
     
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