Question

An iron boiler of mass 180 kg contains 730 kg of water at 11 ∘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.

How long does it take for the water to reach the boiling point from 11 ∘C∘C?

How long does it take for the water to all have changed to steam from 11 ∘C∘C?

Answer 1

(a) As mentioned in the problem,

Specific heat of water, s = 4186 J/kg*C = 4.186 kJ/kg°C

Initial temperature of water, T1 = 11 deg C

Boiling point of water, T2 = 100 deg C

Mass of water, m = 730 kg

Therefore, quantity of heat required to raise the temperature of water from T1 to T2 is

Q = m*s*(T2 - T1)

= 730*4.186*(100 - 11) = 271964.42 kJ

Specific heat of iron, s1 = 450 J /kg°C = 0.450 kJ/kg°C

Mass of iron boiler, m1 = 180 kg

Therefore, quantity of heat required to raise the temperature of iron boiler from T1 to T2,

Q1 = m1*s1*(T2 - T1)

= 180*0.450*(100 - 11) = 7209 kJ

Total quantity of heat required to reach the boiling point of water = Q + Q1

= 271964.42 kJ + 7209 kJ = 279173.42 kJ

Heater supplies energy at the rate, 58000 kJ/h

Therefore, time taken by the heater to supply 279173.42 kJ of energy is -

T = 279173.42 / 58000 = 4.81 hrs. (Answer)

(b) Now, we want to convert the water to steam at 100°C

Latent heat of steam = 2260 kJ/kg

So, this will require:

730 * 2260 = 1649800 kJ of energy.

The iron boiler will remain at 100°C

Now add up Total heat energy required = 279173.42 + 1649800 = 1928973.42 kJ

Therefore, time required = 1928973.42 / 58000 = 33.26 hrs. (Answer)