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Math Help - Specific heat

  1. #1
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    Specific heat

    How much heat (kJ) is needed to raise the temperature of 100.0 grams of water from 25.0C to 50.0C?

    Q=m x C x change in temp

    Q= ?
    m= 100.0g
    C=?
    tf 25.0 C - Ti 50.0 C

    trouble setting this up

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  2. #2
    Senior Member jakncoke's Avatar
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    Re: Specific heat

     Q = 4.8 \frac{J}{g C} * 100 \frac{g}{ml} * 25 C

    about 12 KJ
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  3. #3
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    Re: Specific heat

    I came up with q= 100.0 x 4.18 x 25.0
    10450.
    10.5 kj
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  4. #4
    Senior Member jakncoke's Avatar
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    Re: Specific heat

    Quote Originally Posted by Louisana1 View Post
    I came up with q= 100.0 x 4.18 x 25.0
    10450.
    10.5 kj
    yea u are right, i should have put 4.18
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