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Homework Help: Internal Energy questions

  1. Mar 15, 2004 #1
    I've got a set of questions based on internal energy and specific heat capacities and things. I could just do with a bit of help on some of them. Just a push in the right direction will probably do! Thanks.

    Specific heat capacity of water = 4200 J/kg/K
    Specific heat capacity of ice = 2100 J/kg/K
    Specific heat capacity of lead = 130 J/kg/K
    Specific heat capacity of copper = 390 J/kg/K
    Specific latent heat of fusion of ice = 330 kJ/kg
    Specific latent heat of vaporisation of water = 2.3MJ/kg

    1. Calculate the thermal capacity of an object that contains 950g of copper and 700g of lead.

    2. An electric kettle is rated as 2.5kW. The thermal capacity of the kettle is negligible and then kettle is filled with 1.2kg of water at 15 degrees C.
    Calculate (a) the time taken for the kettle to bring the water to the boil and (b) the mass of the water boiled away in 4 mins.

    3. The mass of 1.0 mole of water is 0.018 kg. Estimate the energy required to remove a molecule of water from the surface of boiling water. (Avogadro constant = 6.02 x 10^23 mol^-1).

    And also if something is as steam and is condensing to water then wuold the specific latent heat of it be the same as the latent heat of vaporisation of water?

    Thanks again!
  2. jcsd
  3. Mar 15, 2004 #2


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    There isn't a whole lot of room to "hint" at these --- you understand the definitions or you don't --- these are "What color was George Washington's white horse?" type questions. One step at a time, and show where you're getting lost.
  4. Mar 15, 2004 #3
    Well for question 1 I don't really understand what it means by thermal capacity but what I think I would do is work out the heat capacity by multiplying the specific heat capacity by the mass and then add the two together.
    For the 2nd question I think I've done part (a) although I'm not sure that it is correct. This is what I did:
    2.5 kW = 2500 J/s
    dQ = mc(dT)
    = 1.2 x 4200 x 85
    = 428400
    428400/2500 = 171.36 secs
    Part (b) I don't really know where to start with

    The 3rd question I am very confused about and again don't really know where to start.
  5. Mar 15, 2004 #4


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    1) Yup.

    2) Good so far. b) Step at a time: decide whether 4 min. is total time, or time elapsed after you get to boiling temp.; then, 4min. or 4 min. - your 171 sec. times burner power equals total energy you've put into vaporizing water.

    3) How much water are you evaporating/vaporizing? 1/A moles?

    Enthalpy of vaporization is minus enthalpy of condensation, or vice versa.
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