Question

a hot iron of mass 0.5 kg is dropped into a container filled with 5 kg of water.The container is made of 0.3 kg of iron initially at 25 celcius .The final equilibrium temperature is 30 celcius .Find the initial temperature of the hot iron

Answer 1

$$m_ic_i(T-T_e)=(m_ic_i+m_wc_w)(T_e-T_0)\\ (440)(0.5)(T-30)\\=((440)(0.3)+(4200)(5))(30-25)\\ T=510\degree C$$