Question

An iron boiler of mass 180 kg contains 690 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⋅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 23 ∘C?

Express your answer using two significant figures.

Part B

How long does it take for the water to all have changed to steam from 23 ∘C?

Express your answer using two significant figures.

Answer 1

apply heat needed to Boil water from 23 to 100 deg

as Q = mcDT

where m is mass of water

c is specific heat

DT is change of tmep = 100-23 = 77 deg C

so

Qw = 690 * 4186 * 77

Qw = 223 MJ of energy

for Iron,

heat needed si Qi = mcDT

Qi = 180 * 450 * 77

Qi = 6.23 MJ

to convert water into Steam,

Latent heat of water = ML = 690 * 2250 = 1.552 MJ

iRon Boiler remains at 100 deg C

so

total eenrgy Q =Qi +Qw + Ql

Q = 223 + 1.552 + 6.23

Q = 230.78 MJ

now use Power P = Energy/time

so

time t = E/P

t = 58000/ 230.78 = 251.32 Hrs total time taken

-----------

time taken by water =58000/223 = 0.260 hrs

time taken by Iron = 58000/1.552 = 37.3 hrs