Chemistry

A small pebble is heated and placed in a foam cup calorimeter containing water at 25 °C. The water reaches a maximum temperature of 27 °C. If the pebble released 532.1 J of heat to the water, what mass (in g) of water was in the calorimeter?

can you show me what numbers you plug where, im confused.

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asked by Rachel
  1. change of water temp = +2 deg C
    heat energy into water = Cwater*m *2 deg

    heat energy out of rock = 532.1 J
    so
    531 J = Cwater * m * 2

    Cwater is specific heat of water in Joules/(gram deg C)

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    posted by Damon
  2. Specific heat water - 4.187 kJ/kgK

    = 4,187 J/gC

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    posted by Damon
  3. It would have helped if you had re-posted the question; however, I found it and here is the response I made.

    heat lost by pebble + heat gained by water = 0
    heat lost by pebble = 532.1 J
    heat gained by water = mass H2O x specific heat H2O x (Tfinal-Tinitial)
    Set q = heat gained by water and solve for mass H2O.
    What's the heat lost by pebble or gained by water. That's 532.1 J.
    mass H2O = x
    specific heat H2O = 4.194 J/g*C
    Tfinal = 27 C
    Tinitial = 25 C
    So
    532.1 = [mass H2O x 4.184 x (27-25)]

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    posted by DrBob222
  4. Specific heat is 4.184 and not 4.194. I typed it right in the equation but wrong above that.

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    posted by DrBob222
  5. I also typed it wrong 4.184 not 4,184

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    posted by Damon

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