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Thread: specific heat capacity problem

  1. #1 specific heat capacity problem 
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    Kelly prepares a cup of juice by mixing juice powder and 0.3kg of hot water at 68 degrees celcius. Find the minimum amount of ice at 0 degrees celcius that should be added to the juice in order to cool the juice to 15 degrees celcius. ( give c (juice) = 4000 J kg^-1 c^-1 )

    4000 x 0.3 x ( 68-15 ) = Energy lost by juice
    63600 = Energy lost by juice

    Energy lost = energy gained.

    so 334000 x m + 2260000 x 15 x m = 63600
    34234000m = 63600
    m=0.001857802kg

    now this is a multiple choice question and options are as follows:
    A 0.16kg
    B 0.19kg
    C 0.53kg
    D 1.00kg

    My answer is clearly wrong, so can someone tell me how to do this.


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  3. #2 Re: specific heat capacity problem 
    Forum Ph.D. william's Avatar
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    Quote Originally Posted by chocolate123
    Kelly prepares a cup of juice by mixing juice powder and 0.3kg of hot water at 68 degrees celcius. Find the minimum amount of ice at 0 degrees celcius that should be added to the juice in order to cool the juice to 15 degrees celcius. ( give c (juice) = 4000 J kg^-1 c^-1 )

    4000 x 0.3 x ( 68-15 ) = Energy lost by juice
    63600 = Energy lost by juice

    Energy lost = energy gained.

    so 334000 x m + 2260000 x 15 x m = 63600
    34234000m = 63600
    m=0.001857802kg

    now this is a multiple choice question and options are as follows:
    A 0.16kg
    B 0.19kg
    C 0.53kg
    D 1.00kg

    My answer is clearly wrong, so can someone tell me how to do this.

    Hey there chocolate,
    The part in red is the culprit. The specific heat of water is c = 4186 J/kg degree(C).

    You should find that the answer is "A".

    Cheers,
    william


    "... the polhode rolls without slipping on the herpolhode lying in the invariable plane."
    ~Footnote in Goldstein's Mechanics, 3rd ed. p. 202
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  4. #3  
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    Quote Originally Posted by chocolate123
    Kelly prepares a cup of juice by mixing juice powder and 0.3kg of hot water at 68 degrees celcius. Find the minimum amount of ice at 0 degrees celcius that should be added to the juice in order to cool the juice to 15 degrees celcius. ( give c (juice) = 4000 J kg^-1 c^-1 )

    4000 x 0.3 x ( 68-15 ) = Energy lost by juice
    63600 = Energy lost by juice

    Energy lost = energy gained.

    so 334000 x m + 2260000 x 15 x m = 63600
    34234000m = 63600
    m=0.001857802kg

    now this is a multiple choice question and options are as follows:
    A 0.16kg
    B 0.19kg
    C 0.53kg
    D 1.00kg

    My answer is clearly wrong, so can someone tell me how to do this.

    You have used 4.00 as the required energy to raise a CC of water by 1 deg celcius, fair enough, but are you sure option D is 1KG and not 1.06Kg?

    You have 63600j to heat x amount of ice to 15 deg at 4.0 calories/cc/deg c
    so it's 63600/60 = 1060grams

    If you take into account latent heat (320j/cc) then it's

    63600/320+(15*4) = 167gms

    Will say's A
    I say D cos I don't think you need worry about latent heat (unless they discussed it with you..)
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  5. #4  
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    D is definitely 1.00kg. So I think the answer's A, since this book did talk about latent heat. I don't get what you mean by 4 calories, doesnt that mean 16.736J ? I dont know how that came in.
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  6. #5  
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    Quote Originally Posted by chocolate123
    D is definitely 1.00kg. So I think the answer's A, since this book did talk about latent heat. I don't get what you mean by 4 calories, doesnt that mean 16.736J ? I dont know how that came in.
    My reading of the original question was that the value 4.0 had merely been 4.186 rounded down, without a knowledge of the level of the book it can be difficult to judge what level is being taught. Anyway you got there in the end.
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  7. #6  
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    Alright.. for me i used 4200Jkg^-1C^-1 as the specific heat capacity for water

    63600 = m x latent heat for ice + m x 4200 x 15
    63600 = m (334000 + 4200(15) )
    m = 0.16

    I got the same answer as will did
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