Question

n number of liquids of masses m, 2m, 3m.... specific heats s, 2s, 3s... at temperatures t, 2t, 3t...are mixed. resultant temperature of mixture​

Answer 1

Answer:

Explanation:

3n2n+1t

2n(n+1)3(n+1)t

3n(n+1)2(n+1)t

3n(n+1)(2n+1)t

Answer :

C

Solution :

From principle of calorimetry

θ=m1S1θ1+m2S2θ2+....m1S1+m2S2+....

m2=6.25m1,%=m2m1+m2×100

θ=(13+23+33+....+n3)t(12+22+32+....+n2).