Question

An iron boiler of mass 180 kg contains 790 kgof water at 21 ∘C. A heater supplies energy at the rate of 58,000 kJ/h. The specific heat of iron is 450 J/kg⋅C∘, the specific heat of water is 4186 J/kg⋅C∘, the heat of vaporization of water is 2260 kJ/kg⋅C∘. Assume that before the water reaches the boiling point, all the heat energy goes into raising the temperature of the iron or the steam, and none goes to the vaporization of water. After the water starts to boil, all the heat energy goes into boiling the water, and none goes to raising the temperature of the iron or the steam.

(A) How long does it take for the water to reach the boiling point from 21 ∘C?

(B) How long does it take for the water to all have changed to steam from 21 ∘C?

Answer 1

Specific heat water = 4.186 kJ/kg°C
You want to heat 790 = 790kg water 79°C. This will take :
790*79*4.186 = 261.24826 MJ heat

Specific heat of iron = 0.450 kJ/kg°C
You want to heat 180kg = 180kg 79°C This will take:
180*79*0.450 = 6.399 MJ

Time taken to reach 100°C = 261.24826 + 6.399 = 267.64726/52 = 5.15 hours

You now convert the water to steam at 100°C
Latent heat of steam = 2260 kJ/kg This will require:
790*2260 = 1.7854 MJ

The iron boiler will remain at 100°C

Now add up Total heat energy required =

261.24826 MJ+ 6.399 MJ+ 1.7854MJ = 269.4 MJ