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24 July, 14:40

A 237-g piece of molybdenum, initially at 100.0 C, is dropped into 244 g of water at 10.0 C. When the system comes to thermal equilibrium, the temperature is 15.3 C. What is the specific heat capacity of molybdenum?

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  1. 24 July, 15:28
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    The specific heat capacity of molybdenum is

    0.2696 joule/gram

    Explanation:

    Step one

    Given that

    Mass of molybdenum Mm = 237g

    Temperature of molybdenum

    T1=100°c

    Mass of water MW = 244g

    Temperature of water T2 = 10°c

    Final temperature T3 = 15.3°c

    Step two:

    We know that the specific heat capacity of water Cw 4.186 joule/gram

    But Specific heat capacity of molybdenum Cm?

    We know that the quantity of heat transfered can be expressed as

    Q=Mm*Cm (T1-T3) + MW*Cw (T2-T3)

    since the system attained thermal equilibrium then the expression

    Is

    =Mm*Cm (T1-T3) = MW*Cw (T3-T2)

    Substituting our values into the equation we have

    237*Cm (100-15.3) = 244*4.186 (15.310)

    20073.9Cm=5413.3352

    Cm=0.2696 joule/gram

    Specific heat capacity of molybdenum is 0.2696 joule/gram
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