Three liquids with masses m1,m2,m3 are thoroughly mixed, If their specific heats are c1,c2,c3 and their temperatures T1,T2,T3 respectively, then the temperature of the mixture is
A
c1T1+c2T2+c3T3m1c1+m2c2+m3c3
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B
m1c1T1+m2c2T2+m3c3T3m1c1+m2c2+m3c3
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C
m1c1T1+m2c2T2+m3c3T3m1T1+m2T2+m3T3
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D
m1T1+m2T2+m3T3c1T1+c2T2+c3T3
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Solution
The correct option is Bm1c1T1+m2c2T2+m3c3T3m1c1+m2c2+m3c3 Let the final temperature be T∘C.
Total heat supplied by the three liquids in coming down to 0∘C=m1c1T1+m2c2T2+m3c3T3......(i)
Total heat used by three liquids in raising temperature from 0∘CtoT∘C=m1c1T+m2c2T+m2c2T+m3c3T.....(ii)
By equating (i) and (ii) we get (m1c1+m2c2+m2c2+m3c3)T=m1c1T1+m2c2T2+m3c3T3⇒T=m1c1T1+m2c2T2+m3c3T3m1c1+m2c2+m3c3