Exactly 14.9 mL of water at 25 0 degree C is added to a hot iron...

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Exactly 14.9 mL of water at 25 0 degree C is added

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Answer 1

Answer #1

Q = m c ∆T
Q = quantity of heat in joules (J)
m = mass of the substance acting as the environment in
grams (g)
c = specific heat capacity (4.19 for H2O) in J/(g ^oC)
∆T = change in temperature = Tfinal - Tinitial in ^oC

Latent heat of vaporization of water = 2259.2 J/g

Specific Heat Capacity of iron = 0.45 J/(g ^oC

14.9 x (100-25) x 4.189 + 14.9 x 22.59 = 1050 x 0.45 x ∆T

∆T = 0.0671140 ^oC

Temperature Change = 0.0671140 ^oC

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Answer 2

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Exactly 14.9 mL of water at 25 0 degree C is added to a hot iron...

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