How much total heat transfer is necessary to lower the temperature of 0.175 kg of steam from 125.5 °C to −19.5 °C, including the energy for

Question

How much total heat transfer is necessary to lower the temperature of 0.175 kg of steam from 125.5 °C to −19.5 °C, including the energy for phase changes? How much time is required for each stage of this process, assuming a constant 835.0 W rate of heat exchange? Give the times in the order that the stages occur.

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Gerda 3 months 2021-07-28T15:06:32+00:00 1 Answers 5 views 0

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    2021-07-28T15:08:00+00:00

    Answer:

    Explanation:

    The heat required to change the temperature of  steam from 125.5  °C to 100 °C is:

    Q_1 = ms_{steam} (125.5^0C - 100^0C) \\ \\ Q_1 = 0.175 \ kg ( 1520 \ J/kg.K ) (25.5^0 \ C) \\ \\ Q_1 = 6783 \ J

    The heat required to change the steam at 100°C to water at 100°C is;

    Q_2 = mL_v \\ \\ Q_2 = (0.175 \ kg) (2.25*10^6 \ J/kg ) \\ \\ Q_2 = 393750 \ J

    The heat required to change the temperature from 100°C to 0°C is

    Q_3 = ms_{water} (100^) \ C) \\ \\ Q_3 = (0.175\ kg)(4186 \ J/kg.K) (100 ^0c ) \\ \\ Q_3 = 73255 \ J

    The heat required to change the water at 0°C to ice at 0°C  is:

    Q_4 = mL_f \\ \\ Q_4 = (0.175 \ kg)(3.34*10^5 \ J/kg) \\\\ Q_4 = 58450 \ J

    The heat required to change the temperature of ice from 0°Cto -19.5°C is:

    Q_5 = ms _{ice} (100^0 C) \\ \\ Q_5 = (0.175 \ kg)(2090 \ J/kg.K)(19.5^0C)  \\ \\ Q_5 = 7132.125 \ J

    The total heat required to change the steam into ice is:

    Q = Q_1 + Q_2 + Q_3 + Q_4 +Q_5 \\ \\Q = (6788+393750+73255+58450+7132.125)J \\ \\ Q = 539325.125 \ J \\ \\ Q = 5.39*10^5 \ J

    b)

    The time taken to convert steam from 125 °C to 100°C is:

    t_1 = \frac{Q_1}{P} = \frac{6738 \ J}{835 \ W}  = 8.12 \ s

    The time taken to convert steam at  100°C to water at  100°C is:

    t_2 = \frac{Q_2}{P} =\frac{393750}{834} =471.56 \ s

    The time taken to convert water to 100° C to 0° C is:

    t_3 = \frac{Q_3}{P} =\frac{73255}{834} = 87.73 \ s

    The time taken to convert water at 0° to ice at 0° C is :

    t_4 = \frac{Q_4}{P} =\frac{58450}{834} = 70.08  \ s

    The time taken to convert ice from 0° C to -19.5° C is:

    t_5 = \frac{Q_5}{P} =\frac{7132.125}{834} = 8.55  \ s

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