What would be the final temperature when 100 g of 25° C water is mixed with 75 g of 40°C water?

1 Answer
Aug 14, 2016

I got about #31.4^@ "C"#.


This is asking you to look at the heat transfer between a hotter substance and a colder substance, and the hotter transfers to the colder. I would expect the result to be closer to #25^@ "C"# than to #40^@ "C"# because there is more colder water than hotter water.

So, you will have to use the heat flow equation:

#bb(q = pmmcDeltaT = pmmc|T_f - T_i|)#

Keep in mind that #q > 0# if heat is absorbed and #q < 0# if heat is released. So #q < 0# for the hotter water and #q > 0# for the colder water. That is so that:

#q_"hotter" + q_"colder" = 0#, i.e. we have conservation of energy for ideal closed systems.

When we assume that the specific heat capacity #c# does NOT change between #25^@ "C"# and #40^@ "C"# significantly:

#q_"hotter" = -mcDeltaT#

#= -("75 g")("4.184 J/g"""^@"C")|T_f - 40^@ "C"|#

#= -("75 g")("4.184 J/g"""^@"C")(40^@ "C" - T_f)#
(because for the hotter water, #T_f < 40^@ "C"#)

#q_"colder" = mcDeltaT#

#= ("100 g")("4.184 J/g"""^@"C")|T_f - 25^@ "C"|#

In an ideal closed system, like the one we assume we have, #q_"hotter" = -q_"colder"#.

So:

#-("75 g")("4.184 J/g"""^@"C")(40^@ "C" - T_f) + ("100 g")("4.184 J/g"""^@"C")(T_f - 25^@ "C") = 0#

#("75 g")cancel(("4.184 J/g"""^@"C"))(40^@ "C" - T_f) = ("100 g")cancel(("4.184 J/g"""^@"C"))(T_f - 25^@ "C")#

#0.75(40^@ "C" - T_f) = T_f - 25^@ "C"#

#30^@ "C" - 0.75T_f = T_f - 25^@ "C"#

#1.75T_f = 55^@ "C"#

#color(blue)(T_f ~~ 31.4^@ "C")#

This makes sense because thermal equilibrium should be established in a way that balances the heat transfer between the water, and that is when both have reached a temperature in between each other's.