The specific heat capacities of three liquids A, B and C are in the ratio, 1 : 2 : 3 and the masses of the liquids are in the ratio 1 : 1 : 1. The temperatures of the liquids A, B and C are `15" "^(@)C, 30" "^(@)C " and 45 "^(@)C`, respectively. Then matched the resultant temperature of the mixture given in list-II with the corresponding mixture given in list-I.

Options

A. `{:(A, B, C, D), ("(i)", (ii), (iii), (iv)):}`
B. `{:(A, B, C, D), ("(ii)", (i), (v), (iii)):}`
C. `{:(A, B, C, D), ("(i)", (iv), (iii), (ii)):}`
D. `{:(A, B, C, D), ("(iv)", (i), (iii), (ii)):}`

Answer

C

Explanation

For (A),
Let the resultant temperature of mixture A and B is `T_(1)` then, from principle of colorimetry `" "(because " temperature B=30^(@) " and temperature of " A=15^(@))`
So, B loses heat and A gains heat
`" "m_(B)c_(B)(30-T_(1))=m_(A)c_(A)(T_(1)-15)`
So, `m times 2(30-T_(1))=m times 1(T_(1)-15)`
`rArr " "2(30-T_(1))=1(T_(1)-15)`
`rArr " "T_(1)=25^(@)C`
Here, `" "m_(1)=m_(2)=m_(3)=1`
and `" "c_(1):c_(2):c_(3)=1:2:3`
For (B),
Let resultant temperature of mixture B and C is `T_(2)` then,
heat lost by C = heat gained by B
`rArr " "3(45-T_(2))=2(T_(2)-30)`
`rArr " "T_(2)=39^(@)C`
For (C),
Let resultant temperature of mixture C and A is `T_(3)`, then,
heat lost by C = heat gained by A
`rArr " "3(45-T_(3))=(T_(3)-15)`
`rArr " "T_(3)=37.5^(@)C`
Hence, `A to I, B to IV, C to III, D to II`.