5kg of water at 20°C is added to 10kg of water at 60°C. Neglecting heat capacity of vessel and other losses, the resultant temperature will be nearly
If T be the resultant temperature, then heat gained by 5kg (at 20∘C ) water H‌gain ‌=mc(T−20)‌‌(∵ given, m=5kg) ⇒‌‌H‌gain ‌=5c(T−20)...(i) Heat lost by 10kg water at 60∘C is given as H‌loss ‌=10c(60−T) ....(ii) By the principle of calorimetry H‌gain ‌=H‌loss ‌ ⇒‌‌5c(T−20)=10c(60−T) ⇒‌‌T−20=120−2T ⇒‌‌3T=140 T=46.67∘C=47∘C