Question

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

Answer 1

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

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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.