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Change of State and Enthalpy; Limiting Reactants

  1. Sep 18, 2013 #1

    Qube

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    Gold Member

    1. The problem statement, all variables and given/known data
    Problem 1:

    An ice cube is dropped into a cup containing 196 grams of coffee at 97 degrees C. The thermal energy used to melt the ice is 2.1 kJ. What is the final temperature of the coffee? Assume the specific heat of the coffee is the same as water.

    Problem 2:

    6 moles of reactant A are mixed with 12 moles of reactant B to produce 9 moles of product C with 3 moles of reactant B left over after the reaction. What is the balanced chemical equation?


    2. Relevant equations

    q=mcΔt

    balanced equations has lowest whole number coefficients

    3. The attempt at a solution

    Problem 1:

    The solid ice must first reach 0 degrees C. Then it must undergo a phase change at this temperature from a solid to liquid. This requires 2100 Joules of energy as given in the problem. This energy is equivalent to the energy the coffee gives up; this energy comes from the coffee; heat flows from the coffee to the ice cube.

    2100 = 195 grams of coffee * 4.184 * (T - 97)

    T = 94.42 degrees C.

    Problem 2:

    The 3 remaining moles imply that only 9 moles of reactant B are required to react with 6 moles of reactant A to form 9 moles of product C.

    Therefore,

    6A + 9B ---> 9C

    Which reduces to

    3A + 2B ---> 3C.

    Am I right? I believe I am but these problems came without solutions and I'm working these old exam problems in preparation for my chemistry test.
     
  2. jcsd
  3. Sep 18, 2013 #2
    Looks OK to me. Just a small error though, it should 97-T instead of T-97. The temperature of coffee decreases.
     
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