Question

A 35.7 gram sample of iron (heat capacity 0.45 g/J°C) was heated to 99.10 °C and placed into a coffee cup calorimeter containing 42.92 grams of water initially at 15.15 °C. What will the final temperature of the system be? (Specific heat of water is 4.184 J/g°C).

Please show work.

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
^{o}C)

∆T = change in temperature = Tfinal - Tinitial in ^{o}C

The heat lost by Iron will equal to heat gained by water

35.7 x 0.45 x (99.10-T_{final} ) = 42.92 x 4.18 x
(T_{final} - 15.15)

(99.10-T_{final} ) = 11.1674
x (T_{final} - 15.15)

(99.10-T_{final} ) = 11.1674 T_{final} -
169.187

99.10 + 169.187 = 12.1674 T_{final}

T_{final} = 268.287 /12.1674 = 22.05

**The final temperature of the system is 22.05
°C**

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