Question

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=

Answer 1

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