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Thermodynamics: Calorimeter problem

  1. Apr 9, 2012 #1
    1. The problem statement, all variables and given/known data
    A 50.0g copper calorimeter contains 250g of water at 20 deg celsius. How much steam must be condensed into the water if the final temperature of the system is to reach 50 deg celsius?


    2. Relevant equations
    Can someone explain how to find the Ls? I was told that just plugging in the latent heat of steam (2.26*10^6) is incorrect.


    3. The attempt at a solution

    Qsteam = Qcalorimeter + Qwater
    ms * Ls = (mc * cc * dT) + (mw * cw * dT)
    ms * Ls = (0.05kg * 0.0924cal/g * (50deg-20deg)) + (.25kg * 1cal/g *(50deg-20deg))
    ms * Ls = 7638.6cal
     
    Last edited: Apr 9, 2012
  2. jcsd
  3. Apr 9, 2012 #2

    Andrew Mason

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    You have to assume that the steam is at 100°C.

    You have the right idea in setting up an equation that equates the heat flow into the 250 g of water and 50 g calorimeter to the heat flow out of an unknown mass of steam. But the heat flow from the steam occurs in two stages: 1. the release of the latent heat of condensation and 2. the heat lost in reducing its temperature from 100C to 50C.

    AM
     
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