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Homework Help: Change in internal energy when water is heated from 0 to 4c

  1. Apr 9, 2017 #1
    1. The problem statement, all variables and given/known data: Find the change in internal energy of 2kg water as it is heated from 0°C to 4°C. The specific heat capacity of water is 4200J/Kg and its densities at 0°C and 4°C are 999.9 kg/m3 and 1000kg/m3 respectively.
    Atm pressure=105Pa

    2. Relevant equations:ΔU= Q-W

    3. The attempt at a solution:
    ΔW = PΔV

    ΔV= M(1/D2-1/D1)
    Plugging the values we get W=-0.02(Volume is decreasing...so definitel W should be negative)
    = (2)(4200)(4)=33600
    ΔU= 33600- (-0.02)
    = 33600 + 0.02 J
    But the answer is given
    33600- 0.02
  2. jcsd
  3. Apr 9, 2017 #2


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    Your answer of (33600 + 0.02) J looks correct to me.
  4. Apr 10, 2017 #3
    then it must be a printing error...right?
  5. Apr 10, 2017 #4


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    Since you are only given the heat capacity to 2 sig figs, correcting the answer by 0.02 seems pointless. But perhaps it's worth demonstrating that it's insignificant?
  6. Apr 10, 2017 #5
    I'm sorry I didnt get you.
  7. Apr 10, 2017 #6


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    Unless otherwise indicated, 4200 means "between 4150 and 4250". Which means Q is between 33200 and 34000 J. With this imprecision, it is meaningless to talk of a correction of 0.02 J.
  8. Apr 10, 2017 #7
    oh..ok..I thought you were giving a hint on how to approach the sum
  9. Apr 13, 2017 #8
    My take on this problem is different. If the initial state is 2 kg water at 0 C and 1 atm, and the final state is 2 kg of water of 4 C and 1 atm, then the path between these two states is irrelevant (since U is a function only of state). Assuming that the specific heat capacity given in the problem statement is ##C_p##, the change in enthalpy between the two states is given by:$$\Delta H=mC_p\Delta T$$The change in internal energy between the two states follows from the definition of the change in enthalpy: $$\Delta H=\Delta U+\Delta (PV)=\Delta U+P_{atm}\Delta V$$So, combining these equations, we get $$\Delta U=mC_p\Delta T -P_{atm}\Delta V$$My point is, there is no need to directly apply the first law of thermodynamics to solve this problem.
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