Question

A 155g piece of iron (Cp = 25.09J/mol. degree Celsius) was heated to a temperature of 52 degrees Celsius and then placed in contact with a 270g piece of copper at 10 degrees Celsius (Cp = 25.46 J/mol. degrees Celsius). What was the final temperature of the two pieces of metal degrees Celsius?

Answer 1

1st find the number of moles of each metals

Molar mass of Fe = 55.85 g/mol


mass(Fe)= 155 g

number of mol of Fe,
n = mass of Fe/molar mass of Fe
=(155.0 g)/(55.85 g/mol)
= 2.775 mol

Molar mass of Cu = 63.55 g/mol


mass(Cu)= 270 g

number of mol of Cu,
n = mass of Cu/molar mass of Cu
=(270.0 g)/(63.55 g/mol)
= 4.249 mol

Let us denote Copper by symbol 1 and iron by symbol 2
m1 = 4.249 mol
T1 = 10.0 oC
C1 = 25.46 mol/goC
m2 = 2.775 mol
T2 = 52.0 oC
C2 = 25.09 J/mol.oC
T = to be calculated

Let the final temperature be T oC
use:
heat lost by 2 = heat gained by 1
m2*C2*(T2-T) = m1*C1*(T-T1)
2.775*25.09*(52.0-T) = 4.249*25.46*(T-10.0)
69.6247*(52.0-T) = 108.1795*(T-10.0)
3620.487 - 69.6247*T = 108.1795*T - 1081.7954
T= 26.4 oC

Answer: 26.4 oC