(5 ~kg ) of water at (20^ C ) is added to (10 ~kg ) of water at (60^ C ). Neglecting heat capacity of vessel and other losses, the resultant temperature will be nearly

Step-by-step solution

If (T ) be the resultant temperature, then heat gained by (5 ~kg ) (at (20^ C ) ) water ( aligned & H_ gain =m c(T-20) ( given, m=5 ~kg ) & H_ gain =5 c(T-20) (i) aligned ) Heat lost by (10 ~kg ) water at (60^ C ) is given as (H_ loss =10 c(60-T) ) ...(ii) By the principle of calorimetry ( array ll & H_ gain =H_ loss & 5 c(T-20)=10 c(60-T) & T-20=120-2 T & 3 T=140 & T=46.67^ C 47^ C array )