given:
m1 = 75.00 g
c1 = ?
t1 = 52.00°C
m2=225.0 g
c2=2.045 J/g-°C
t2=07.50°C
tf=9.08°C
Using
ΣmcΔt=0
75.00*c1*(9.08-52.00)+225.0(2.045)(9.08-7.50)=0
Solve for c1 to get c1=0.2258 J/g-°C
A 75.00-g sample of an unknown metal is heated to 52.00°C. It is then placed in a coffee cup calorimeter filled with water. The calorimeter and the water inside have a combined mass of 225.0 g and an overall specific heat of 2.045 J/(gi°C). The initial temperature of the calorimeter is 7.50°C. If the system reaches a final temperature of 9.08°C when the metal is added, what is the specific heat of the unknown metal?
1 answer
1 answer