Chemistry Help ?

A 177‐g sample of barium at 186.5 °C is placed in 130.8 mL of water at 32 °C. What is the final temperature (in °C) of the water? Assume that no heat is lost to or gained from the surroundings. The specific heat capacity of barium = 0.201 J/g⋅°C, the specific heat capacity of water = 4.184 J/g⋅°C, and the density of water, ρwater = 0.997 g/mL.

an explanation on how to get an answer rather than just the answer is more appreciated, but both would be great, thanks in advance!

2 Answers

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  • 3 weeks ago

    this seems to be a simple thermodynamics problem, but barium reacts with water

    Ba + 2H₂O --> Ba(OH)₂ + H₂

    and I have no idea how much heat that releases.

  • Anonymous
    3 weeks ago

    The approach to all of these is to remember that the energy given off by the hot materials = energy gained by the cold materials.  At equilibrium, everything is at the same temperature.

    There's no phase change in this example so the amount of energy transferred from the hot is:

    energy = mass of hot * specific heat of hot * temperature change of the hot

    For the cold, you use the same equation since again, we have no phase change

    energy = mass of cold * specific heat of cold * temperature change of cold.

    the energy is the same for both so we can equation the two equations

    mass of hot * specific heat of hot * temperature change of the hot

     = mass of cold * specific heat of cold * temperature change of cold

    Substituting and letting T be the final temperature which we don't know

    177 g * 0.201 J/gC * (186.5C - T) = 130.8 mL * 0.997 g/mL * 4.184 J/gC * (T - 32C)

    Now you just have to crunch the numbers and find T

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