pls help asap?

Calculate the thermal energy required to change 15 Kg of ice at – 20.0 degrees Celsius to liquid at 20 degrees Celsius. *

Q = 6,907,800 J

Q = 4,507,600 J

Q = 5,000,700 J

Q = 9,907,500 J

How much thermal energy must be removed from 2 kg of water at 0 degrees Celsius to make ice cubes at 0 degrees Celsius? *

Q = 670,000 J

Q = 690,000 J

Q = 580,000 J

Q = 100.670 J

10 kg of water at 90 degrees Celsius have to be converted to steam at 100 degrees Celsius. Determine the quantity of heat required(Specific latent heat of vaporization of water = 22.6 x 10 exp -5 J kg-1)(Specific heat capacity of water = 4186 J kg-1 K-1) *

Q = 117,018,600 J

Q = 20,018,000 J

Q = 23,018,600 J

Q = 25,005,456 J

How much heat must be added to 4.5 Kg of copper to change it from a solid at 25900 K to a liquid at 25900 K? *

Q = 180,000 J

Q = 815,000 J

Q = 362,000 J

Q = 931,500 J

one of the following is not a characteristics of the equilibrium vapor-pressure curve for water? *

It determines the boiling point of a liquid.

The equilibrium vapor pressure of water increases with increasing temperature.

At the boiling point of water, the vapor pressure is equal to atmospheric pressure.

It determines the freezing point of a liquid

2 Answers

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  • 2 months ago
    Favorite Answer

    I will do one problem as an example.

    10 kg of water at 90 degrees Celsius have to be converted to steam

    at 100 degrees Celsius. Determine the quantity of heat required.

    (Specific latent heat of vaporization of water = 22.6 x 10 exp -5* J kg-1)

    (Specific heat capacity of water = 4186 J kg-1 K-1**)

    *The above figure is not correct. For water, Lv = 2.26E+06 J/kg.

    **ΔK = Δ°C

    m = mass of water = 10 kg

    T0 = initial temperature = 90 °C

    T = final temperature = 100 °C

    Δt = change in temperature = T - T0 = 10 °C

    Lv = specific latent heat of vaporization of water = 2.26E+06 J/kg

    c = specific heat capacity of water = 4,186 J/kg∙°C

    Q = quantity of heat required = to be determined

    Q = mLv + mcΔt

    Q = (10 kg)(2.26E+06 J/kg) + (10 kg)(4,186 L/kg∙°C)(10°C)

    Q = 2.30E+07 J = 23,018,600 J

  • Amy
    Lv 7
    2 months ago

    energy required to heat 15kg of ice by 20 degrees

    +

    energy required to melt 15 Kg of ice

    +

    energy required to heat 15kg of water by 20 degrees

    Look up the heat of fusion of water and the specific heat capacity of liquid water and ice.

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