Question

A piece of iron weighing 20.0 g at a temperature of 95.0ºC was placed in 100.0 g of Water at 25.0ºC. Assuming that no heat is lost to the surroundings, what is the resulting temperature of the iron and water?

Answer 1

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

m1 = 100.0 g

T1 = 25.0 oC

C1 = 4.184 J/goC

m2 = 20.0 g

T2 = 95.0 oC

C2 = 0.45 J/goC

T = to be calculated

Let the final temperature be T oC

we have below equation to be used:

heat lost by 2 = heat gained by 1

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

20.0*0.45*(95.0-T) = 100.0*4.184*(T-25.0)

9*(95.0-T) = 418.4*(T-25.0)

855 - 9*T = 418.4*T - 10460

T= 26.5 oC

Answer: 26.5 oC