Question

**A piece of iron weighing 20.0 g at a temperature of
95.0****º****C was placed in 100.0 g of
Water at 25.0****º****C. 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

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