In a particular lab, a cube of ice (Tice = -5.5˚C) is taken and dropped into a calorimeter cup (98g) partially filled with 326 g of water (W

Question

In a particular lab, a cube of ice (Tice = -5.5˚C) is taken and dropped into a calorimeter cup (98g) partially filled with 326 g of water (Water = 20˚C). The cup was at the same initial temperature as the water and is perfectly insulating. The final temperature of the system is 15˚C. What was the mass of ice added?

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Mộc Miên 5 months 2021-08-30T06:30:40+00:00 1 Answers 2 views 0

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    2021-08-30T06:31:46+00:00

    Answer:

    The mass of the ice added = 16.71 g

    Explanation:

    The heat gained by the ice is equal to the heat lost by the calorimeter cup and the water in the cup.

    But for this question, the cup is said to be perfectly insulated, hence, there is no loss of heat from the calorimeter cup.

    Heat gained by the ice = Heat lost by the 326 g of water.

    Let the mass of ice be m

    The heat gained by the ice = (Heat gained by ice in temperature from -5.5°C to 0°C) + (Heat used by the ice to melt at 0°C) + (Heat required for the melted ice to rise in temperature from 0°C to 15°C)

    Heat gained by ice in temperature from -5.5°C to 0°C = mCΔT

    m = unknown mass of ice

    C = Specific Heat capacity of ice = 2.108 J/g°C

    ΔT = change in temperature = 0 – (-5.5) = 5.5°C

    Heat gained by ice in temperature from -5.5°C to 0°C = m×2.108×5.5 = (11.594m) J

    Heat used by the ice to melt at 0°C = mL

    m = unknown mass of ice

    L = Latent Heat of fusion of ice to water = 334 J/g

    Heat used by the ice to melt at 0°C = m×334 = (334m) J

    Heat required for the melted ice or water now, to rise in temperature from 0°C to 15°C = mCΔT

    m = unknown mass of water (which was ice)

    C = Specific Heat capacity of water = 4.186 J/g°C

    ΔT = change in temperature = 15 – 0 = 15°C

    Heat required for the melted ice or water now, to rise in temperature from 0°C to 15°C = m×4.186×15 = (62.79m) J

    Total heat gained by the ice = 11.594m + 334m + 62.79m = (408.384m) J

    Heat lost by the water in the calorimeter cup = MCΔT

    M = mass of water in the calorimeter cup = 326 g

    C = specific heat capacity of water = 4.186 J/g°C

    ΔT = change in temperature = 20 – 15 = 5°C

    Heat lost by the water in the calorimeter cup = 326×4.186×5 = 6,823.18 J

    Heat gained by the ice = Heat lost by the 326 g of water.

    408.384m = 6,823.18

    m = (6,823.18/408.384)

    m = 16.71 g

    Hope this Helps!!!

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