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

Question

Wyatt Atkins · Accepted Answer

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.

Therefore, you must apply the equation for heat flow:

#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"#.

Thus:

#-("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, once both water temperatures have reached a point where they are in between one another, thermal equilibrium should be established in a way that balances the heat transfer between the two.

Henry Antrim · Answer

undefined To find the final temperature when mixing two substances, you can use the principle of energy conservation, specifically the principle of heat exchange. This can be expressed as:

$m_1 \cdot c_1 \cdot (T_f - T_1) = -m_2 \cdot c_2 \cdot (T_f - T_2)$

Where:
- $m_1$ and $m_2$ are the masses of the substances
- $c_1$ and $c_2$ are the specific heat capacities of the substances
- $T_1$ and $T_2$ are the initial temperatures of the substances
- $T_f$ is the final temperature of the mixture

Given:
- $m_1 = 100$ g
- $m_2 = 75$ g
- $T_1 = 25$ °C
- $T_2 = 40$ °C
- $c_1 = c_2 = c$ (assuming water, specific heat capacity is constant)

Plugging in the values:

$100 \cdot c \cdot (T_f - 25) = -75 \cdot c \cdot (T_f - 40)$

Solving for $T_f$:

$100 \cdot c \cdot T_f - 2500c = -75 \cdot c \cdot T_f + 3000c$

$175 \cdot c \cdot T_f = 5500c$

$T_f = \frac{5500}{175}$

$T_f = 31.43$ °C

So, the final temperature when 100 g of 25°C water is mixed with 75 g of 40°C water is approximately 31.43°C.