A blacksmith cools a 1.12kg-chunk of iron, initially at a temperature of 650.0 o

Question

A blacksmith cools a 1.12kg-chunk of iron, initially at a temperature of 650.0oC, by trickling 15.6oC water over it. All the water boils away, and the iron ends up at a temperature of 120.0oC. How much water did the blacksmith trickle over the iron?
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Note: The text book gives the specific heat capacity of iron as 0.47J/kgoC and water as 4.19J/kgoC

Also: The answer given was 0.107kg for the mass of water

The way I tried to answer this problem was Qiron=Qwater (insofar as the heat gained by the water equalled the heat lost by the iron). However, I kept getting answers around 0.790kg for the mass of water. Do I need to include something about the energy that is used up in the phase change of the water into steam? Please explain this for me. Thank you!

Explanation / Answer

m1 c1 dT = m2 cw dt

1.12 *0.47 * [ 650 -120 ] = m * 4.19 * [ 100 - 15.6 ] + m * 2257 + m * 1.996 * 20
m =0.1052 kg 

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