n liquids with masses m, 2m, 3m,..., nm having specific heats S, 2S, 3S,..., nS and temperatures t, 2t, 3t,..., nt are mixed together. The final (equilibrium) temperature of the mixture is:

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

Correct answer: 3n(n+1)/(2(2n+1)) t

Solution

The numerator is mSt*sum(k³) = mSt*[n(n+1)/2]² and the denominator is mS*sum(k²) = mS*n(n+1)(2n+1)/6; their ratio simplifies to 3n(n+1)/(2(2n+1)) times t.

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