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16 January, 05:40

If a 100.0 g sample of water at 6.7°C is added to a 100.0 g sample of water at 57.0°C, determine the final temperature of the water. Assume no heat is lost to the surroundings.

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Answers (2)
  1. 16 January, 06:29
    0
    The final temperature will be 31.85 °C

    Explanation:

    Step 1: Data given

    Mass of water = 100.0 grams

    Temperature of water = 6.7 °C

    Mass of the other sample water = 100.0 grams

    Temperature = 57.0 °C

    Step 2: Calculate the final temperature

    Energy Lost by the hot water = Energy Gain by the cold water

    mass of hot water*c * (Temp. of hot water-x) = mass of cold water*c * (x-Temp. of cold water)

    Energy Lost by the hot water=Energy Gain by the cold water

    mass of hot water*c * (Temp. of hot water-x) = mass of cold water*c * (x-Temp. of cold water)

    100*c * (57-x) = 100*c * (x-6.7)

    100 * (57-x) = 100 * (x-6.7)

    5700 - 100x = 100x - 670

    6370 = 200x

    x = 31.85

    The final temperature will be 31.85 °C
  2. 16 January, 08:20
    0
    The answer to your question is Te = 31.85 °C

    Explanation:

    Data

    mass 1 = 100 g

    temperature 1 = 6.7°C

    mass 2 = 100 g

    temperature 2 = 57°C

    equilibrium temperature = ?

    C = 1

    Formula

    mCΔT = - mCΔT

    Substitution

    100 (Te - 6.7) = - 100 (Te - 57)

    Simplification

    100 Te - 670 = - 100 Te + 5700

    100Te + 100Te = 5700 + 670

    200Te = 6370

    Te = 6370 / 200

    Result

    Te = 31.85 °C
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