Question

An aluminum block of mass 260g and at T= 89 degrees celcius is placed in a cup holding 320g of water at 22 degreed celcius

The system is allowed to reach thermal equilibrium.

What is the entropy change of the water?

Answer 1

Assuming heat capacity of Al = 0.91 kJ/ kgK , and that of water = 4.186 kJ/ kg K

Given, Mass of Al (m al)=260g = 0.26 kg ; Mass of water(m w) = 320 g =0.320 kg

The aluminium block , initially at Th= 89 degrees C= 362 K , is placed in a cup of water at Tc=22 degrees C= 295 K, and the attain thermal equilibrium.

Let us assume final temperature as Tm degrees C

Assuming no heat loss to the surroundings, we can say that Heat lost by aluminium is equal to the heat gained by water..

Therefore,       \(m_{Al} c_{Al} (T_{h} -T_{m})=m_{w} c_{w} (T_{m} -T_{c})\)   

from here, we get,       \(T_{m}=(m_{Al} c_{Al} T_{h} +m_{w} c_{w} T_{c})/(m_{Al} c_{Al}+m_{w} c_{w})\)   

=> Tm = (0.26*0.91*362 + 0.32*4.186*295)/(0.26*0.91 + 0.32*4.186) = 305.05 K

Now, entropy change for water is given by,

     

C= mass of water x specific heat of water = 1.33952 kJ/k

Tm= 305.05 K , and Tc=295 K

Substituting, we get, Change in Entropy = 1.33952 ln(305.05/295) = 0.0311 kJ/K