A hot lump of 47.6 g of iron at an initial temperature of 50.8 °C is placed in 50.0 mL of H2O initially at 25.0 °C and allowed to reach thermal equilibrium. What is the final temperature of the iron and water given that the specific heat of iron is 0.449 J/(g·°C)? Assume no heat is lost to surroundings.
The heat balance:
-Qhot = Qcold
Qhot = Qiron
Qcold = Qwater
substitute
Q = m*C*(Tf-Ti)
where m mass, C specific heat, T temperatures
Qiron = m iron * Ciron * (Tf-Tiron)
Q water= m water* Cwater* (Tf-Twater)
then
-m iron * Ciron * (Tf-Tiron) = m water* Cwater* (Tf-Twater)
assume mass of water = 1 g/mL for 50 mL --> 50 g
-47.6 * 0.449 * (Tf - 50.8) = 50 * 4.184*(Tf-25)
Tf - 50.8 = -9.7883*Tf + 25*9.7883
Tf(1+9.7883) = 50.8 + 25*9.7883
T F= (50.8 + 25*9.7883) /(10.7883)
Tf = 27.39
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