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A solution is made by mixing 296.0 mL of ethanol initially at 13.7 ∘C with 296.0 mL of water initially at 22.5 ∘C .?

A solution is made by mixing 296.0mL of ethanol initially at 13.7∘C with 296.0mL of water initially at 22.5∘C. What is the final temperature of the solution assuming that no heat is lost? The density of ethanol is 0.789g/mL and the density of water is 1.00g/mL. The specific heat of ethanol is 2.46J/g⋅°C and the specific heat of water is 4.184J/g⋅°C. Please show me how to do this

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  • 2 months ago
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    From law of conservation of energy,

    heat gained by ethanol = heat lost by water

    m1*s1*DT1 = m2*s2*DT2

    m1 = mass of ethanol = 190.149 g

    (Given that density of ethanol is 0.789 g/mL

    the mass of 296.0 mL of ethanol is 296*0.789 = 233.544 gms)

    s1 = specific heat of ethanol= 2.46 j/g.c

    DT = (t-13.7) oC

    m2 = mass of water = 296 g

    s2 = specific heat of water = 4.184 j/g.c

    DT2 = (22.5-t)

    233.544*2.46*(t-13.7) = 296*4.184*(22.5-t)

    The final temperature of the mixture is 19.71 oC

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  • Anonymous
    2 months ago

    Same way you do any of these problems.  You remember that the heat lost by the hot material = heat gained by the cold.  Let T be the final temperature for the water and the ethanol.

     

    (296.0 mL * 0.789g/mL * 2.46J/g⋅°C * (T -13.7C) = 296.0 mL * 1.00g/mL * 4.184J/g⋅°C * (22.5C - T)

     

    Solve for T

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