Question

The temperature of equal masses of three different liquids

Answer 1

The temperature of liquid when A & C are mixed is 20.25° C.

Explanation:

Given data: The temperature of three of three different liquids of equal masses are  

Ta = 12° C

Tb = 19° C

Tc = 28° C

The temperature of the mixture of A & B, Tab = 16° C

The temperature of the mixture of B & C, Tbc = 23° C

To find: The temperature of the mixture of A & C, “Tac”

Let the specific heat of the three liquids A, B & C be “Sa”, “Sb” & “Sc” respectively.

Based on the law of calorimetry, we can write  

mSa(Tab - Ta) = mSb(Tb - Tab)

or, Sa(16 - 12) = Sb(19 - 16)

or, 4Sa = 3Sb

or, Sa/Sb = ¾ ….. (i)

And,

mSb(Tbc - Tb) = mSc(Tc - Tbc)

or, Sb(23 - 19) = Sc(28 - 23)

or, 4Sb = 5Sc

or, Sb/Sc = 5/4 ….. (ii)

From (i) & (ii), we get

Sa:Sb:Sc = 15:20:16

∴ Sa = (15/16)Sc ….. (iii)

Therefore,  

The temperature of mixture A & C is given by

mSa(Tac - Ta) = mSc(Tc - Tac)

putting the value of Sa from (iii)

or, 15/16Sc(Tac – 12) = Sc(28 - Tac)

or, 15(Tac – 12) = 16(28 - Tac)

or, 15Tac – 180 = 448 – 16Tac

or, 31Tac = 628

or, Tac = 628/31 = 20.25° C