Calculate the total energy, in kilojoules, that is needed to turn a 46 g block of ice at -25 degrees C into water vapor at 100 degrees

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Calculate the total energy, in kilojoules, that is needed to turn a 46 g block
of ice at -25 degrees C into water vapor at 100 degrees C.

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  1. Answer: The amount of heat absorbed is 141.004 kJ.

    Explanation:

    In order to calculate the amount of heat released while converting given amount of steam (gaseous state) to ice (solid state), few processes are involved:

    (1): [tex]H_2O (s) (-25^oC, 248K) \rightleftharpoons H_2O(s) (0^oC,273K)[/tex]

    (2): [tex]H_2O (s) (0^oC, 273K) \rightleftharpoons H_2O(l) (0^oC,273K)[/tex]  

    (3): [tex]H_2O (l) (0^oC, 273K) \rightleftharpoons H_2O(l) (100^oC,373K)[/tex]

    (4): [tex]H_2O (l) (100^oC, 373K) \rightleftharpoons H_2O(g) (100^oC,373K)[/tex]

    Calculating the heat absorbed for the process having the same temperature:

    [tex]q=m\times \Delta H_{(f , v)}[/tex]       ……(i)

    where,

    q is the amount of heat absorbed, m is the mass of sample and [tex]\Delta H_{(f , v)}[/tex] is the enthalpy of fusion or vaporization

    Calculating the heat released for the process having different temperature:

    [tex]q=m\times C_{s,l}\times (T_2-T_1)[/tex]      ……(ii)

    where,

    [tex]C_{s,l}[/tex] = specific heat of solid or liquid

    [tex]T_2\text{ and }T_1[/tex] are final and initial temperatures respectively

    • For process 1:

    We are given:

    [tex]m=46g\\C=2.108J/g^oC\\T_2=0^oC\\T_1=-25^oC[/tex]

    Putting values in equation (i), we get:

    [tex]q_1=46g\times 2.108J/g^oC\times (0-(-25))\\\\q_1=2424.2J[/tex]

    • For process 2:

    We are given:

    [tex]m=46g\\\Delta H_{fusion}=334J/g[/tex]

    Putting values in equation (i), we get:

    [tex]q_2=46g\times 334J/g\\\\q_2=15364J[/tex]

    • For process 3:

    We are given:

    [tex]m=46g\\C=4.186J/g^oC\\T_2=100^oC\\T_1=0^oC[/tex]

    Putting values in equation (i), we get:

    [tex]q_3=46g\times 4.186J/g^oC\times (100-0)\\\\q_3=19255.6J[/tex]

    • For process 4:

    We are given:

    [tex]m=46g\\\Delta H_{vap}=2260J/g[/tex]

    Putting values in equation (i), we get:

    [tex]q_4=46g\times 2260J/g\\\\q_4=103960J[/tex]

    Calculating the total amount of heat released:

    [tex]Q=q_1+q_2+q_3+q_4[/tex]

    [tex]Q=[(2424.2)+(15364)+(19255.6)+(103960)][/tex]

    [tex]Q=141003.8J=141.004kJ[/tex]                  (Conversion factor: 1 kJ = 1000J)

    Hence, the amount of heat absorbed is 141.004 kJ.

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