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

##### 1 Answer

I got about

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

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

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

Keep in mind that **absorbed** and **released**. So *hotter* water and *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**

#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,

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.