613 kg of water at room temperature (20^oC) is heated in a 127 kg iron boiler. I

Question

613 kg of water at room temperature (20^oC) is heated in a 127 kg iron boiler. If energy is supplied at a rate of 48,000 kJ/h, how long will it take for the water to completely boil away?

Iron Lf = 289 J/g

cIron = 0.450 J/gC

MPIron = 1538^oC

29 hours 35 hours 33 hours 45 hours

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Explanation / Answer

To raise the temperratue of water to 100 deg C, the heat required,

= m*C*dT = 613*4186*(100-20) = 2.053*10^8 J

To completely evaporate the water heat required = 613*2257*10^3 = 1.38*10^9 J

So, total heat required, Hw = 2.053*10^8 + 1.38*10^9

= 1.585*10^9 J

Now, heat required to raise 127 kg iron to 100 deg C

Hi = M*C*dT = 127*450*(100-20) = 4.57*10^6 J

So, total heat required, Ht = Hw + Hi = 1.585*10^9 +  4.57*10^6 = 1.589*10^9 J

So, time required, t = 1.589*10^9/(48000*10^3)

= 33.1 hours <--------answer

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