Question

Two portions of the same liquid are mixed together. One has a mass of 100.0 g and a temperature of 43 °C. The other has a mass of 25 g and a temperature of 21 °C. Ignoring the container in which they are mixed and any heat lost from that container, determine the final temperature T_f at equilibrium.

Answer 1

Masses m = 100 g

           M = 25 g

Initial temperatures t = 43 ^o C

                             t '= 21 ^o C

Final temeprature T = ?

Heat lost by m = heat gain by M

mc (t-T ) = Mc(T-t ')

m(43-T) = M(T-21)

100(43-T)=25(T-21)

  4 (43-T) = T-21

172 - 4T = T-21

          5T = 193

            T = 38.6 ^o C