Question

In: Physics

How much total heat transfer is necessary to lower the temperature of 0.275 kg of steam...

How much total heat transfer is necessary to lower the temperature of 0.275 kg of steam from 145.5 °C to −15.5 °C, including the energy for phase changes?

total heat transfer:

J

How much time is required for each stage of this process, assuming a constant 815.0 W rate of heat exchange? Give the times in the order that the stages occur.

t1=

t2=

t3=

t4=

t5=

Solutions

Expert Solution


let
Q1 is the heat lost during temperature decreases from 145 to 100 C
Q2 is the heat lost during phase change from vapour to water
Q3 is the heat lost during water temperature decreases from 100 C to 0C
Q4 is the heat lost during water becomes ice
Q5 is heat heat lost during temperature of ice decreases from 0 to -15.5 C

we know,
C_steam = 1996 J/(kg C)
Lv = 2.26*10^6 J/kg
C_water = 4186 J(kg C)
Lf = 3.33*10^5 J/kg
C_Ice = 2100 J/(kg C)


Q1 = m*C_ateam*(145 - 100)
= 0.275*1996*45
= 24700 J

Q2 = m*Lv
= 0.275*2.26*10^6
= 6.215*10^5 J

Q3 = m*C_water*(100 - 0)

= 0.275*4186*100

= 115115 J

Q4 = m*Lf

= 0.275*3.33*10^5

= 9.1575*10^4 J

Q5 = m*C_Ice*(0 - (-15.5))

= 0.275*2100*15.5

= 8951 J

Total heat transfer,

Q = Q1 + Q2 + Q3 + Q4 + Q5

= 24700 + 6.215*10^5 + 115115 + 9.1575*10^4 + 8951

= 861841 J <<<<<<<<-----------------Answer

part b)

t1 = Q1/P = 24700/815 = 30.3 s
t2 = 6.215*10^5/815 = 763 s
t3 = 115115/815 = 141 s
t4 = 9.1575*10^4/815 = 112 s
t5 = 8951/815 = 11 s


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