A blacksmith cools a 1.22-kg chunk of iron, initially at a temperature of 650.0C

Question

A blacksmith cools a 1.22-kg chunk of iron, initially at a temperature of 650.0C, by trickling 15.2 Cwater over it. All the water boils away, and the iron ends up at a temperature of 120.0C.

Part A

How much water did the blacksmith trickle over the iron?

Express your answer with the appropriate units.

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A blacksmith cools a 1.22-kg chunk of iron, initially at a temperature of 650.0C, by trickling 15.2 Cwater over it. All the water boils away, and the iron ends up at a temperature of 120.0C.

Part A

How much water did the blacksmith trickle over the iron?

Express your answer with the appropriate units.

SubmitMy AnswersGive Up

Explanation / Answer

mass of iron = 1.22 kg, T1=650 0C, T2 = 120.0 0C
mass of water= ?, T1 = 15.2 0C , T2 = 100 0C

Heat lost by iron is
   Q = m*c*delta T
       = 1.22*460*(650-120)

       =1.22*460*(530)
       = 297436 J

heat gained by water is Q2 = m c delta T
               = m *4186* (100 - 15.2)
              = m*4186*84.8
                           = 354972.8m J
    we know that heat of vaporization o f water is 2264000 J/kg

   sphesific heat of water 4146 J/kg

heat gained by water being vaporized is 2264000m - 354972.8m = 1909027.2 m
   
   1909027.2m J = 297436 J   ==>      m = 0.15580 kg of water

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