Constants An iron boiler of mass 180 kg contains 750 kg of water at 23 C. A heat

Question

Constants An iron boiler of mass 180 kg contains 750 kg of water at 23 C. A heater supplies energy at the rate of 58,000 kJ/h. The specific heat of iron is 450 J/kg. Co, the specific heat of water is 4186 J/kg C, the heat of vaporization of water is 2260 kJ/kg . Co. 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 di

Explanation / Answer

(a) Given that -

specific heat water = 4186 J/kg°C = 4.186 kJ/kg°C

mass of water = 750 kg

So, amount of heat required by this water to reach the temperature at 100 deg C

Q1 = 750*(100-23)*4.186 = 241741.5 kJ

Specific heat of iron = 0.450 kJ/kg°C
You want to heat 180 kg iron through 77°C

This will take:
Q2 = 180*77*0.450 = 6237 kJ

Total amount of heat required = Q1 + Q2 = 241741.5 + 6237 = 247978.5 kJ

Therefore, time taken by the heater = 247978.5 / 58000 = 4.275 hours.

(b) Now, you convert the water to steam at 100°C
Latent heat of steam = 2260 kJ/kg

So, amount of heat required for this conversion, Q3 = 750*2260 = 1695000 kJ

The iron boiler will remain at 100°C

So, total amount of heat required = Q1 + Q2 + Q3 = 241741.5 + 6237 + 1695000 = 1942978.5 kJ

Therefore, the time required = 1942978.5 / 58000 = 33.50 hours.

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