A blacksmith cools a 1.18-kg chunk of iron, initially at a temperature of 650.0C

Question

A blacksmith cools a 1.18-kg chunk of iron, initially at a temperature of 650.0C, by trickling 14.2 C water 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.

A blacksmith cools a 1.18-kg chunk of iron, initially at a temperature of 650.0C, by trickling 14.2 C water 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.

Explanation / Answer

here,

mass of iron , m1 = 1.18 kg

initial temprature , Ti = 650 degree C

final temprature , Tf = 120 degree C

let the mass of water be m2

using conservation of heat energy

heat energy lost by iron = heat energy gained by water

m1 * Ci * ( Ti - Tf) = m2 * Cw * ( (100 - 14.2) + Lv)

1.18 * 450 * (650 - 120 ) = m2 * 4186 * ( 85.8 ) + m2 * 2250000

solving for m2

m2 = 0.107 kg

the mass of water is 0.107 kg

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