Question

How much total heat transfer is necessary to lower the temperature of 0.295 kg of steam from 145.5 C to -23.5 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.

T1 =

T2 =

T3 =

T4 =

T5 =

Answer 1

Heat released is given by:

Q = Q1 + Q2 + Q3 + Q4 + Q5

Q1 = Heat released from 145.5 C steam to 100 C steam = Ms*Cs*dT1

Q2 = Heat released for phase change from steam to water = Ms*Lv

Q3 = Heat released from 100 C water to 0 C water = Ms*Cw*dT2

Q4 = Heat released for phase change from water to ice = Ms*Lf

Q5 = Heat released from 0 C water to -23.5 C ice = Ms*Ci*dT3

here, Ms = mass of steam = 0.295 kg

dT1 = 45.5 degC

dT2 = 100 degC

dT3 = 23.5 degC

Using above given values:

Q1 = 0.295*2010*45.5 = 26979.225 J

Q2 = 0.295*2.25*10^6 = 663750 J

Q3 = 0.295*4186*100 = 123487 J

Q4 = 0.295*3.34*10^5 = 98530 J

Q5 = 0.295*2090*23.5 = 14488.925 J

So, total heat transfer required = Q1 + Q2 + Q3 + Q4 + Q5 = 26979.225 + 663750 + 123487 + 98530 + 14488.925 = 927235.15 J

Now, Energy(Q) = power(P)*time(T)

then, time(T) = Q/P

given, P = 835.0 W

So, T1 = Q1/P = 26979.225/835.0 = 32.31 s

T2 = Q2/P = 663750 /835.0 = 794.91 s

T3 = Q3/P = 123487 /835.0 = 147.89 s

T4 = Q4/P = 98530 /835.0 = 118 s

T5 = Q5/P = 14488.925 /835.0 = 17.35 s

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