Three liquids A, B and C of specific heats `1cal//g-^@C, 0.5 cal//g-^@C and 0.25cal//g-^@C` are at temperatures `20^@C, 40^@C and 60^@C` respectively. Find temperature in equilibrium if they are mixed together. Their masses are equal.

Answer

A, C, D

Explanation

Let `theta (gt60^@)` is the mixture temperature in equilibrium.
Heat given = Heat taken
`0 = m(1)(theta -20) + m(0.5)(theta -40)`
`+ m (0.25)(theta -60)`
Solving the equation, we get
`theta= 31.43^@C`