Question

Three liquids are at temperatures of 9^◦C, 23^◦C, and 37^◦C, respectively. Equal masses of the first two liquids are mixed, and the equilibrium temperature is 17^◦C. Equal masses of the second and third are then mixed, and the equilibrium temperature is 32.4^◦C.

Find the equilibrium temperature when equal masses of the first and third are mixed.

Answer in units of ^◦C.

Answer 1

Let Cp1 , CP2 and Cp3 are specific heats of liquid-1, liquid-2 and liquid-3

if equal masses of liquid-1 and liquid-2 are mixed to get equilibrium temperature 17o C, then we have

thermal energy gain by liquid-1 = thermal energy loss from liquid-2

Cp1 (17-9) = Cp2 (23-17)   i.e., Cp1 = 0.75 Cp2 ..............(1)

if equal masses of liquid-2 and liquid-3 are mixed to get equilibrium temperature 32.4 oC, then we have

thermal energy gain by liquid-2 = thermal energy loss from liquid-3

Cp2 (32.4-23) = Cp3 (37-32.4)   i.e., Cp3 = 2.043 Cp2 ..............(2)

if equal masses of liquid-1 and liquid-3 are mixed to get equilibrium temperature T , then we have

thermal energy gain by liquid-1 = thermal energy loss from liquid-3

Cp1 (T - 9) = Cp3 (37-T)   ..............(3)

By substituting Cp1 and Cp3 from eqn.(1) and eqn.(2), we get

0.75 (T-9) = 2.043 ( 37 - T).................(4)

By solving eqn.(4) , we get T = 29.48 oC