8.33 kg of steam at temperature of 150 C has 2.14×107 J of heat removed from it.

Question

8.33 kg of steam at temperature of 150 C has 2.14×107 J of heat removed from it. Determine the final temperature and phase of the result once the heat has been removed if the heat is removed at constant pressure during the gas phase.

1.How much heat would need be removed to lower the temperature of the steam to the boiling point? Give your answer as a positive value since you were asked for the amount removed.

2.How much heat would be needed to change the steam into liquid water at the boiling temperature?

3.How much heat would be needed to lower the temperature of the liquid water to the melting point?

4.What is the final temperature of the material in C?

Explanation / Answer

Specific heat of steam, Cs = 1926 J/kg.C

heat required to decrease the temperature of steam to 100 deg C :

= 8.33*1926*(150-100)

= 8.02*10^5 J <--------------answer (1)

Now, the heat removed is more than the amount required to decrease the temperature ot 100 deg C

So, it will decrease to a lower value of temperature.

So, there will be a phase change at 100 deg C to water

So, heat required for whole 8.33 kg of stem to change to water = 8.33*2260000 = 1.88*10^7 J <-------answer(2)

So, total heat lost upto phase change = 1.88*10^7 + 8.02*10^5 = 1.96*10^7 J

After phase change, let the temperature change be T <---- that is T deg less than 100 deg C

total heat lost = 8.33*4190*(100 - T) + 1.96*10^7 = 2.14*10^7

So, T = 48.4 deg <-------- final temperature <-------(answer 4)

heat required to lower the temperature to melting point = 8.33*4190*100 = 3.49*10^6 J <------answer(3)

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