You cool a 130.0 g slug of red-hot iron (temperature 745 ∘C) by dropping it into an insulated cup of negligible mass containing 85.0 g of wa

Question

You cool a 130.0 g slug of red-hot iron (temperature 745 ∘C) by dropping it into an insulated cup of negligible mass containing 85.0 g of water at 20.0 ∘C. Assume no heat exchange with the surroundings. How do you do this?

Part A What is the final temperature of the water?
Part B What is the final mass of the iron and the remaining water?

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Lệ Thu 5 years 2021-08-29T23:26:01+00:00 1 Answers 9 views 0

Answers ( )

  1. Helga
    0
    2021-08-29T23:27:34+00:00
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    Answer:

    a.T_f=100.00\textdegree C\\\\b. 211\ grams\\

    Explanation:

    a.

    -The specific heat of iron is 0.45 J /g/K

    -Heat released by red hot iron to cool to 100°C

    =130g \times 0.45J/g/K\times 645\textdegree C\\\\=37732.5J

    #The heat required by water to heat up to 100°C :

    = 85g \times 4.2 \times (100-20)\textdegree C\\\\ = 28560 J

    #You notice that this heat is less that heat supplied by the iron slug so the equilibrium temperature will be 100 ° C.

    #Let m be the grams of water vaporized during the prices:

    H_v=m \times 540x4.2\\ \\ = 2268m J

    #Heat required to warm the water of 85 g to 100 °C is calculated as:

    H_w=85X4.2 X 80\\ \\= 28560 J\\

    #For an equilibrium:

    Heat Lost= Heat Gained

    37732.5J=28560J+2268m \ J\\\\m=4 \ g

    Hence, the amount of water vaporized is 4 grams

    b. Final mass of iron and remaining water is calculated as;

    M_b=M_b-m\\\\=81g+130g-4g\\\\=211g\\

    Hence, the final mass of the iron and the remaining water is 211 grams.

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