what is the final temperature of 250g of water whose initial Temperature is 25 C if 80-g of aluminum initially at 70 C is dropped into the water? The specific heat of aluminum is 0.215 cal/C g how do you do it? our teacher did not show us how and it is not in the bookGiven Data :-- Mass : Aluminium m(a) = 80 g, Water m(w) = 250 g Initial Temperature : Aluminium t₁ = 70 C and Water t₁\' = 25 C Final Temperature of both ( aluminium and water ) = T C (assume) Since Aluminium is initially at a temperature higher than that of water, it (aluminium) will lose heat to water . Heat lost by Al = Q = mass of Aluminium x sp.heat capacity of Aluminium x Temp. change = Q = 80 x 0.215 x ( 70 - T ) cal Heat gained by water = Q\' = mass of water x sp.heat capacity of water x Temp. change = Q\' = 250 x 1 x ( T - 25 ) According to principle of calorimetery . = Heat lost by aluminium = heat gained by water. = 250 x 1 x ( T - 25 ) = 80 x 0.215 x ( 70 - T ) = T = 27.9 C .... ( Rounded to one decimal place )...... Answer Answer .Other related question