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.
Q = m c ∆T
Q = quantity of heat in joules (J)
m = mass of the substance acting as the environment in
c = specific heat capacity (4.19 for H2O) in J/(g oC)
∆T = change in temperature = Tfinal - Tinitial in oC
The heat lost by Iron will equal to heat gained by water
35.7 x 0.45 x (99.10-Tfinal ) = 42.92 x 4.18 x (Tfinal - 15.15)
(99.10-Tfinal ) = 11.1674 x (Tfinal - 15.15)
(99.10-Tfinal ) = 11.1674 Tfinal - 169.187
99.10 + 169.187 = 12.1674 Tfinal
Tfinal = 268.287 /12.1674 = 22.05
The final temperature of the system is 22.05 °C
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