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A 32.8 g iron rod, initially at 23.0 ∘C, is submerged into an unknown mass of water at 64.0 ∘C, in an insulated contain...

A 32.8 g iron rod, initially at 23.0 ∘C, is submerged into an unknown mass of water at 64.0 ∘C, in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 59.0 ∘C.

What is the mass of the water?

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Answer #1

Heat lost by water = -(heat gained by iron rod)

For iron rod,

mass = m = 32.8 g

Specific heat of iron = c = 0.449 J/goC

\DeltaT = Tf-Ti = 59.0-23.0 = 36.0 oC

Q = mc\DeltaT = (32.8)(0.449)(36.0) = 530.18 J

For water,

mass = m = ?

Specific heat of water = c = 4.18 J/goC

\DeltaT = Tf-Ti = 59.0-64.0 = -5.0 oC

Q =  -(heat gained by iron rod) = -530.18 J

Q = mc\DeltaT

-530.18 = (m)*(4.18)(-5.0)

m = 25.37 g

Mass of water = 25.37 g

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