Question

A hot lump of 43.0 g of iron at an initial temperature of 62.6 °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 equilibrium temperature attained by the system

Heat gained by water = Heat lost by iron

                    mcdt = m'cdt'                  

Where

m = mass of water at 25.0oC = volume x density = 50.0 mL x 1.0g/mL = 50.0g

c = specific heat capacity of water = 4.184 J/g-oC

dt = change in temperature of water = (t-25.0 ) oC

m' = mass of iron = 43.0 g

c' = specific heat capacity of iron = 0.449 J/g-oC

dt' = change in temperature of iron = (62.6-t) oC

Plug the values we get 50x4.184x (t-25.0 ) = 43.0x0.449x(62.6-t)

                                      209.2 x (t-25.0 ) = 19.307x(62.6-t)

                                     10.83 x (t-25.0 ) = (62.6-t)

                                                t =50.5 oC

So the final temperature of the substances is 50.5 oC