6th grade

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100 milliliters of 10 degrees celsius water plus 100 milliliters of 90 degrees celsius water, what is the final temperature?

  • 6th grade -

    simplified energy balance:
    Q,absorbed + Q,lost = 0
    where Q = mc(T2-T1)
    in which Q=heat in J, m=mass in kg, c=specific heat capacity (which has a definite value for various substances) in J/(kg*C), T=temperature in Kelvin
    *T2=final temp (the same for both and is usually the unknown)
    *T1=initial temp

    since in this problem, it involves only one type of substance (which is water), the c will be canceled:
    Q,absorbed + Q,lost = 0
    m1*c*(T2-T1,a) + m2*c*(T2-T1,b) = 0
    m1*(T2-T1,a) + m2*(T2-T1,b) = 0
    ..from here, recall that density is mass per unit volume or:
    d=m/V
    density of water is approximately equal to 1 g/mL, thus,
    m1 = 100 g
    m2 = 100 g
    since they are the same, we cancel them from the equation:
    m1*(T2-T1) + m2*(T2-T1) = 0
    (T2-T1,a) + (T2-T1,b) = 0
    solving for T2:
    T2 - 90 + T2 - 10 = 0
    2(T2) = 100
    T2 = 50 degrees Celsius

    *note that change in T is involved thus Celsius and Kelvin unit can be used interchangeably

    so there,, sorry for long explanation~

  • 6th grade -

    t2=50 degrees celisius

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