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The question is 50g of steam at 165 degree is added to 200g of ice at -45 degree, what will be the final temperature of the system?

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  • 1 decade ago
    Favorite Answer

    you have to use the formula q=mc(final temp - initial temp)

    The steam will lose heat (q) and therefore will be negative.

    -q = mc(final temp - initial temp)

    The ice will gain heat and will still be q = mc (final - initial).

    The heat loss for the steam is equal to the heat gained from the ice so you can set the two formulas equal together, so...

    - mc (final - initial) = mc (final - initial)

    ^^^^^ steam ^^^^^^^ ice

    So now we can put in the numbers.

    The specific heat (c) of steam is 1.87 J/g Celsius

    The specific heat of ice is 2.06 J/g Celsius

    The final temperature of the two will be the same.

    -50 g * 1.87 J/gC (final-165C) = 200 g * 2.06 J/gC (final - (-45C))

    Plug some numbers into your calculator and you get

    -93.5 J/C (final-165C) = 412 J/C (final + 45C)

    Divide both sides by 95.5 J/C and you get

    final - 165C = -4.406 (final + 45 C)

    Then distribute the -4.406 and you get

    final -165 C = -4.046*final - 198.29 C

    Combine the like terms and you get

    5.406*final = -33.29 C

    Divide both sides by 5.406 and you get your answer of

    final temperature = -6.158 C

    Gotta Love Thermo

  • Oracle
    Lv 4
    1 decade ago

    After what period of time?

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