A liquid of mass M and specific heat S is at temperature 2t. If another liquid of thermal capacity 1.5 times at a temperature of t/3 of same mass is added to it, the resultant temperature will be
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a
(4/3)t
b
t
c
t/2
d
2t/3
answer is B.
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Detailed Solution
t0=M1S1t1+M2S2t2M1S1+M2S2 Given :M1=M, S1=S, t1=2t M2S2=32MS,t2=t3 ∴ t0=2MSt+32×t3MSMS+32MS=t