Question

A piece of Fe metal with a mass of 0.350 kg at a temperature of 300 °C is added to 1.00 kg of H2O(1) at a temperature of 20 °C. What is the final equilibrium temperature of the H:0(0) and the Fe? The specific heat of the Fe metal is 0.502 J/g.°C and the specific heat of Hh0 is 4.18 Jlg. C.

Answer 1

Formula is as bellow

Q = mCp(T₁ - T_f) where m = mass, Cp = Specific heat, T₁ = Initial temprature while T_f = final temprature, Q = heat transfer. Now the heat lost by iron = heat gain by water

So the final eqution will be

0.35*0.502*(300-T_f) = 1*4.18*(T_f-20)

Or, 0.175*(300-T_f) = 4.18*(T_f-20)

Or, 300 + 477.71 = 24.885T_f

Or,  T_f = 31.25^o C

Hence the final equibrium temprature of both of them will be 31.25^o C