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.

Answer 1

Step-by-step 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 = (Q_1)/(P) = (6738 \ J)/(835 \ W) = 8.12 \ s`

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

`t_2 = (Q_2)/(P) =(393750)/(834) =471.56 \ s`

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

`t_3 = (Q_3)/(P) =(73255)/(834) = 87.73 \ s`

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

`t_4 = (Q_4)/(P) =(58450)/(834) = 70.08 \ s`

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

`t_5 = (Q_5)/(P) =(7132.125)/(834) = 8.55 \ s`