A 2.45 kg aluminium pan at 155 C is plunged into 3.58 kg of water. If the entropy change of the system is 162 J/k, what is the initial temperature of the water?
Q = mcΔT ΔS=mcln(T_2/T_1) Q_water + Q_Aluminium = 0
c water = 4184 J/kg*K c aluminium = 900 J/kgK
3. Attempted Solutions
T_i Aluminum = 155
T_f is the same for the water and aluminium
T_i where T_i is the initial temperature of the water.
162 = (3.58)(4184)(ln(T_f/T_i) + (2.45)(900)ln((T-f)/(428))
(3.54)(4814)(T_f-T_i) + (2.45)(900)(T_f - 428)= 0
After solving this system of equations multiple times I can't seem to get the correct answer. The answer is 16 degrees C but I am not sure how to get there.