Question

A hot lump of 32.9 g of iron at an initial temperature of 55.9 °C is placed in 50.0 mL of H2O initially at 25.0 °C and allowed to reach thermal equilibrium. What is the final temperature of the iron and water given that the specific heat of iron is 0.449 J/(g·°C)? Assume no heat is lost to surroundings.

Answer 1

Let t be the common temperature attained by the system

Heat lost by iron = heat gained by water

               mcdt = m'c'dt'

Where

m = mass of iron = 32.9 g

c = specific heat capacity of iron = 0.449 J/(goC)

dt = change in temperature of iron = 55.9-t

m' = mass of water = 50.0mL x ( 1.00 g/mL) = 50.0 g

c' = specific heat capacity of water = 4.18 J / g ^oC

dt' = change in temperature of water = t - 25.0

Plug the values we get  

32.9x0.449x(55.9-t) = 50.0x4.18x(t-25.0)

                55.9 - t = 14.15x(t - 25.0)

                           = 14.15 t - 353.7

                15.15 t = 409.6

                         t = 27.0 ^oC

Therefore the final temperature of the iron and water is 27.0 ^oC