If you combine 250.0 mL of water at 25°C and 120.0 mL of water at 95°C, what is the final temperature of the mixture?

1 Answer
Jul 4, 2016

Answer:

#T_f^oC = 47.07^oC#

Explanation:

We can use a heat equilibrium equation to find this value.

#DeltaH = m(T_f-T_i)C_p#

#DeltaH =# Change in Heat
#m =# Mass
#T=# Tempurature
#C_p=# Specific Heat

#DeltaH# cold #H_2O# = -#DeltaH# hot #H_2O#

#m(T_f-T_i)C_p = -[m(T_f-T_i)C_p]#

Since the density of water is #1 g/(mL)# the volume of water is also equal to the mass of the water in grams.

Cold #H_2O#

#m = 250.0g#
#T_f= ?#
#T_i = 25^oC#
#C_p = 4.18 J/(g^oC)#

Cold #H_2O#

#m = 120.0g#
#T_f= ?#
#T_i = 95^oC#
#C_p = 4.18 J/(g^oC)#

#m(T_f-T_i)C_p = -[m(T_f-T_i)C_p]#

#250.0g(T_f-25^oC)4.18 J/(g^oC) = -[120.0g(T_f-95^oC)4.18 J/(g^oC)]#

#1,045T_^oC - 26,125J/(g^oC) = -501.6T_f^oC + 47,652J/(g^oC)#

# (cancel(1,546.6)T_f^oC)/(cancel(1,546.6)) = (73,777)/(1,546.6)#

#T_f^oC = 47.07^oC#