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 ^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-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