Using formula Q= mC(Tf-Ti)…. I don’t understand this?

If 70 Btu of heat energy are added to 2.5 lb of water at 35F, what will be the final temperature of the water? Use formula Q = mC(Tf – Ti), and solve for Tf

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3 Answers

  • Q = heat

    m = mass

    C = specific heat

    Tf = final temperature

    Ti = initial temperature

    So...

    70 Btu = 2.5 lbs(covert to mass) * specific heat of water * (Tf - 35)

    If we assume the 2.5 lbs of water is lbm (mass), then:

    70 = 2.5(1)(Tf - 35)

    70 = 2.5(Tf) - 2.5(35)

    70 = 2.5Tf - 87.5

    2.5Tf = 157.5

    Tf = 157.5/2.5

    Tf = 63 degF

    Best of luck!

  • Q = mC(Tf-Ti)

    Q = 70 Btu

    m = 2.5 lb

    C of water = 1 Btu /lb °F

    Ti = 35° F

    70 = 2.5 (Tf - 35)

    Tf -35 = 70/2.5 = 28

    Tf = 28 + 35 = 63° F

  • Just divide both parts of the equation by mC.

    Q/mC = mC/mC * (Tf-Ti)

    Q/mC = Tf - Ti.

    Q/mC + Ti = Tf.

    Just put the values in the equation and that's it. If you don't know C for water, it's a "constant" called specific heat. Right now, you just consider C to be a constant value... look it up in google, and you have everything just put it into your calculator and that's it.

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