Question

A 32.5 g iron rod, initially at 22.6 ∘C, is submerged into an unknown mass of water at 63.1 ∘C, in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 59.2 ∘C.

What is the mass of the water?

Express your answer to two significant figures and include the appropriate units.

Answer 1

Let us denote iron by symbol 1 and H2O by symbol 2

m1 = 32.5 g

T1 = 22.6 oC

specific heat capacity of iron ,

C1 = 0.45 J/goC

m2 = to be calculated

T2 = 63.1 oC

C2 = 4.184 J/goC

T = 59.2 J/goC

we have below equation to be used:

heat lost by 2 = heat gained by 1

m2*C2*(T2-T) = m1*C1*(T-T1)

m2*4.184*(63.1-59.2) = 32.5*0.45*(59.2-22.6)

m2*16.3176 = 535.275

m2*4.184*(63.1-59.2) = 32.5*0.45*(59.2-22.6)

m2= 33 g

Answer: 33 g