Let Q1,Q2,Q3 be the temperatures of container C1,C2 and C3 respectively. Using principle of calorimetry in container C1, we have (θ1−60)=2ms(60−θ) ⇒θ1−60=120−2θ ⇒θ1=180−2θ ......(i) For container C2 ms(θ2−30)=2ms(30−θ) ⇒θ2=90−2θ3 .......(ii) For container C3 2ms(θ1−60)=ms(60−θ) ⇒2θ1−120=60−θ ⇒2θ1+θ=180 .......(iii) Also, θ1+θ2+θ3=3θ ........(iv) Adding (i), (ii) and (iii) 3θ1+3θ2+3θ3=450 ⇒θ1+θ2+θ3=150 ⇒3θ=150⇒θ=50°C