Question

A hot lump of 45.0 g of iron at an initial temperature of 68.2 degrees celsius is placed in 50.0 mL of H2O initially at 25.0 degrees celsius and allowed to reach thermal equillibrium. What is the final temperature of the iron and water given that the specific heat of iron is 0.449 J/(G x degrees celsius)? Assume no heat is lost to surroundings.

Answer 1

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

m1 = 50.0 g

T1 = 25.0 oC

C1 = 4.184 J/goC

m2 = 45.0 g

T2 = 68.2 oC

C2 = 0.449 J/goC

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)

45.0*0.449*(68.2-T) = 50.0*4.184*(T-25.0)

20.205*(68.2-T) = 209.2*(T-25.0)

1377.981 - 20.205*T = 209.2*T - 5230

T= 28.8 oC

Answer: 28.8 oC