Question

# Three liquids are at temperatures of 9◦C, 23◦C, and 37◦C, respectively. Equal masses of the first...

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

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

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