The following physical constants are for water, H2O. The specific heat capacity of the solid = 2.09 J/g oC The specific heat ca

Question

The following physical constants are for water, H2O.
The specific heat capacity of the solid = 2.09 J/g oC
The specific heat capacity of the liquid = 4.18 J/g oC
The specific heat capacity of the vapor = 2.09 J/g oC
∆Hfus = 6.02 kJ/mol; ∆Hvap = 40.7 kJ/mol Freezing point = 0.0oC; Boiling point = 100.0oC
How much heat(in kJ) is required to warm 10.0 grams of ice at -5.0oC to a temperature of 70.0oC?

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Latifah 6 months 2021-07-13T19:26:19+00:00 1 Answers 0 views 0

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    2021-07-13T19:28:05+00:00

    Answer:

    Q\approx6.4~kJ

    Explanation:

    Quantity of heat required by 10 gram of ice initially warm it from -5°C to 0°C:

    Q_1=m.C_s.\Delta T

    here;

    mass, m = 10 g

    specific heat capacity of ice, C_s=2.09~J.g^{-1}.^{\circ}C^{-1}

    change in temperature, \Delta T=(5-0)=5^{o}C

    Q_1=10\times2.09\times 5

    Q_1=104.5~J

    Amount of heat required to melt the ice at 0°C:

    Q_2=m.\Delta H_{fus}

    where, \Delta H_{fus}=6020~J/mol

    we know that no. of moles is = (wt. in gram) \div (molecular mass)

    Q_2=\frac{10}{18} \times 6020

    Q_2=3344.44~J

    Now, the heat required to bring the water to 70°C from 0°C:

    Q_3=m.C_L.\Delta T

    specific heat of water, C_L=4.18~J/g/^oC

    change in temperature, \Delta T=(70-0)=70^oC

    Q_3=10\times 4.18\times 70

    Q_3=2926~J

    Therefore the total heat required to warm 10.0 grams of ice at -5.0°C to a temperature of 70.0°C:

    Q=Q_1+Q_2+Q_3

    Q=104.5+3344.44+2926

    Q=6374.94~J

    Q\approx6.4~kJ

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