An insulating cup contains 200 grams of water at 25 ∘C. Some ice cubes at 0 ∘C is placed in the water. The system comes to equilibrium with

Question

An insulating cup contains 200 grams of water at 25 ∘C. Some ice cubes at 0 ∘C is placed in the water. The system comes to equilibrium with a final temperature of 12∘C. How much ice in grams was added to the water?
The specific heat of ice is 2090 J/(kg ∘C), the specific heat of water is 4186 J/(kg ∘C), latent heat of the ice to water transition is 3.33 x10^5 J/kg

in progress 0
Khoii Minh 4 years 2021-07-27T12:54:49+00:00 1 Answers 18 views 0

Answers ( )

  1. ngocdiep
    0
    2021-07-27T12:56:05+00:00
    Reply

    Answer:

    The amount of ice added in gram is 32.77g

    Explanation:

    This problem bothers on the heat capacity of materials

    Given data

    Mass of water Mw= 200g

    Temperature of water θw= 25°c

    Temperature of ice θice= 0°c

    Equilibrium Temperature θe= 12°c

    Mass of ice Mi=???

    The specific heat of ice Ci= 2090 J/(kg ∘C)

    specific heat of water Cw = 4186 J/(kg ∘C)

    latent heat of the ice to water transition Li= 3.33 x10^5 J/kg

    heat heat loss by water = heat gained by ice

    N/B let us understand something, heat gained by ice is in two phases

    Heat require to melt ice at 0°C to water at 0°C

    And the heat required to take water from 0°C to equilibrium temperature

    Hence

    MwCwΔθ=MiLi +MiCiΔθ

    Substituting our data we have

    200*4186*(25-12)=Mi*3.3×10^5+

    Mi*2090(12-0)

    837200*13=Mi*3.3×10^5+Mi*2090

    10883600=332090Mi

    Mi=10883600/332090

    Mi= 32.77g

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )