chemistry

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A 115 g piece of metal, initially at 60.0 °C, is submerged into 100.0 g of water at 25.0 °C. At thermal equilibrium, this mixture had a final temperature of 27.0 °C. The specific heat capacity of water is 4.18 (J/g°C). Use this information to determine the specific heat capacity of the metal (calculate your answer to 3 sig figs).

  • chemistry -

    The heat change for both objects will be equal, so you need two equations and they need to be set to each other. The equation to use is the following:

    q=mcT∆

    Conditions for 1
    m1=mass1=115g
    c1=specific heat1=?
    T∆1=25.0°C-27.0ºC=-2.0Cº

    Conditions for 2:
    m2=mass2=100.0g
    c2=specific heat2=4.18 (J/g°C)
    T∆2=25.0°C-60.ºC=-30.0ºC



    m1c1∆T1=m2c2∆T2
    (115g)c1(-2.0Cº)=(100.0g)(4.18 (J/g°C))(-30.0ºC)

    Solve for c1,

    c1=[(100.0g)(4.18 (J/g°C))(-30.0ºC)]/[(115g)(-2.0Cº)]

    c1=(-1.25 x 10^4 J/)(-2.3 x 10^2 gºC)

    c1=54.3 J/g°C

    ****I think; I havent had to do this type of calculations in years.

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