An iron boiler of mass 180 kg contains 810kg of water at 13?C . A heater supplie

Question

An iron boiler of mass 180 kg contains 810kg of water at 13?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.

Part A

How long does it take for the water to reach the boiling point from 13?C ?

Part B

How long does it take for the water to all have changed to steam from 13?C ?

Explanation / Answer

apply heat needed to Boil water from 13 to 100 deg

as Q = mcDT

where m is mass of water

c is specific heat

DT is change of tmep = 100-13 = 87 deg C

so

Qw = 810 * 4186 * 87

Qw = 295 MJ of energy

for Iron,

heat needed si Qi = mcDT

Qi = 180 * 450 * 87

Qi = 7.047 MJ

to convert water into Steam,

Latent heat of water = ML = 810 * 2250 = 1.822 MJ

iRon Boiler remains at 100 deg C

so

total eenrgy Q =Qi +Qw + Ql

Q = 295 + 1.822 + 7.047

Q = 303.869 MJ

now use Power P = Energy/time

so

time t = E/P

t = 58000/ 303.869    = 190.87 Hrs total time taken

-----------

time taken by water =58000/295 = 0.196 hrs

time taken by Iron = 58000/1.822 = 31.8 hrs

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