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Lecture Demonstrations

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    50915
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    Calorimeter Constant

    To a styrofoam cup calorimeter containing 250 mL of water at 24.0 oC is added a known amount of heat as 250 mL of water from a second styrofoam cup at 32.0 oC. The final temperature, measured by a computer-interfaced thermistor, is 27.7 oC. The computer-generated plot of T vs. time should be projected[1]. What is the calorimeter constant?

    ΔT calorimeter = 27.7 oC – 24.0 oC = 3.7 oC

    ΔT cold = 27.5 oC – 24.0 oC = 3.7 oC

    ΔThot = 27.7 – 32.0 = -4.3 oC

    qhot = m x S.H. x ΔT = 250 g x 4.18 J/goC x -4.3oC = -4494 J

    qcold = m x S.H. x ΔT = 250 x 4.18 x 3.7 oC = 3867 J

    qcalorimeer + qhot + qcold = 0

    qcalorimeter -4494 + 3867 = 0

    qcalorimeter = 627 J.

    C = Qcalorimeter/T = 627 J / 3.7 oC = 169 J/oC

     

    The Ammonium Nitrate "Cold Pack"[2]

    A styrofoam cup calorimeter contains 250 mL of water at 25.0oC. Solid NH4NO3 (5 g ) is added, and the temperature falls to 23.8oC. The computer-generated plot of T vs. time should be projected[3]. The calorimeter is found to absorb 169 J to change its temperature 1 oC, so it is said to have a calorimeter constant of 169 J/oC. What is the enthalpy change for the dissolution reaction?

    qwater = m x S.H. x ΔT = 250 g x 4.18 J/goC x -1.2oC

    = -1254 J

    qcalorim = C x ΔT = 169 x -1.2oC = -203 J

    q tot = -1254 + -203 = -1457 J

    q rxn = +1457 J.

    ΔH rxn = q (kJ) / n (mol)

    n = m / M = 5 g NH4NO3 / 80 g/mol = 0.057 mol

    ΔH rxn = q/n = 1.457 kJ / 0.057 mol = +25.6 kJ/mol

    The reaction is spontaneous even though it is endothermic, because of the large positive entropy change resulting from water association with the separate ions in solution.

     

    References

    1. We used Vernier LoggerPro(R) software
    2. J. Chem. Educ., 2004, 81 (1), p 64A
    3. We used Vernier LoggerPro(R) software

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