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Math Help - phys quiz question please help me, thank you

  1. #1
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    Talking phys quiz question please help me, thank you

    The quantity of heat Q that changes the temperature T of a mass m of a substance is given by Q = mcT, where c is the specific heat capacity of the substance. For example, for H20, c = 1 cal/gC. And for a change of phase the quantity of heat Q that changes the phase of a mass m is Q = mL, where L is the heat of fusion or heat of vaporization of the substance. For example, for H20 the heat of fusion is 80 cal/g or 80 kcal/kg, and the heat of vaporization is 540 cal/g or 540 kcal/kg. Use these relationships to determine the number of calories to change
    (a) 0.6 kg of 0C ice to 0.6 kg 0C ice water
    kcal
    (b) 0.6 kg of 0C ice water to 0.6 kg 100C boiling water
    kcal
    (c) 0.6 kg of 100C boiling water to 0.6 kg l00C steam
    kcal
    (d) 0.6 kg of 0C ice to 0.6 kg 100C steam
    kcal
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  2. #2
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    Quote Originally Posted by johnjohn View Post
    The quantity of heat Q that changes the temperature T of a mass m of a substance is given by Q = mcT, where c is the specific heat capacity of the substance. For example, for H20, c = 1 cal/gC. And for a change of phase the quantity of heat Q that changes the phase of a mass m is Q = mL, where L is the heat of fusion or heat of vaporization of the substance. For example, for H20 the heat of fusion is 80 cal/g or 80 kcal/kg, and the heat of vaporization is 540 cal/g or 540 kcal/kg. Use these relationships to determine the number of calories to change
    (a) 0.6 kg of 0C ice to 0.6 kg 0C ice water
    kcal
    (b) 0.6 kg of 0C ice water to 0.6 kg 100C boiling water
    kcal
    (c) 0.6 kg of 100C boiling water to 0.6 kg l00C steam
    kcal
    (d) 0.6 kg of 0C ice to 0.6 kg 100C steam
    kcal
    Hello,

    you only have to use all the given constants:

    (a): 600\ g \cdot 80\ \frac{cal}{g}=48000\ cal=48\ kcal

    (b): 600 \ g \cdot 1 \ \frac{cal}{g \cdot ^\circ C} \cdot 100^\circ C=60000\ cal=60\ kcal

    (c): 600\ g \cdot 540\ \frac{cal}{g}=324000\ cal=324\ kcal

    (d); Add the results (a)+(b)+(c)

    A personal question: I'm used to measure the difference of temperatures in Kelvin and if I remember correctly the unit cal or kcal is not allowed since 1979
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