Describe what would happen to the temperature of water if a person poured 2 liters of water at 25 degrees C into a container that has 2 liters of water at 75 degrees C. What temperature would the mixture become?

Question

Serenity Allen · Accepted Answer

Conceptually, we should be able to figure this out in less than #2# minutes (i.e. this is a straightforward ACS-style question).

The final temperature for two equal masses of water combined will be halfway between the high and low temperatures.

Mathematically, we assume conservation of thermal energy out from the hot body of water into the cold body of water:

#q_(cold) = mC_PDeltaT_(cold)#

#= -q_(hot) = -mC_PDeltaT_(hot)#,

where #q# is heat flow, #q_(hot) < 0#, and #q_(cold) > 0#. #C_P = "4.184 J/g"^@ "C"# is the specific heat capacity of water at constant pressure (lab bench conditions), and #m# is the mass of the water in #"g"#.

Both bodies of water have the same volume and hence approximately the same mass (more or less...), assuming similar densities at these different temperatures.

If we assume we don't know what the final temperature is, except for that it will be the same for both bodies of water (as required to reach thermal equilibrium), then:

#cancel(mC_P)DeltaT_(cold) = -cancel(mC_P)DeltaT_(hot)#

#=> T_f - T_(i,cold) = -(T_f - T_(i,hot))#

#=> T_f - 25^@ "C" = -(T_f - 75^@ "C")#

#=> 25^@ "C" - T_f = T_f - 75^@ "C"#

#=> 100^@ "C" = 2T_f#

#=># #color(blue)(T_f ~~ 50^@ "C")#

Indeed, #50^@ "C"# is right in the middle of #25^@ "C"# and #75^@ "C"#, as predicted conceptually.

Avery Adams · Answer

undefined The final temperature of the mixture would be an intermediate value between 25 degrees Celsius and 75 degrees Celsius, approaching the initial temperature of the hotter water (75 degrees Celsius). The specific calculation can be done using the principle of conservation of energy and the heat capacity equation.

Caleb Adams · Answer

undefined To determine the final temperature of the water mixture, we can use the principle of conservation of energy, specifically the principle of heat transfer. This principle states that the total heat lost by the hot water (initially at 75°C) will be equal to the total heat gained by the cold water (initially at 25°C) plus any heat absorbed by the container itself.

We can calculate the amount of heat exchanged using the formula:

$$Q = mcΔT$$

where:
- $Q$ is the heat transferred,
- $m$ is the mass of the substance (in this case, water),
- $c$ is the specific heat capacity of water (which is approximately 4.18 J/g°C), and
- $ΔT$ is the change in temperature.

For the hot water:
$$Q_{\text{hot}} = m_{\text{hot}}cΔT_{\text{hot}}$$

For the cold water:
$$Q_{\text{cold}} = m_{\text{cold}}cΔT_{\text{cold}}$$

Since the final temperature will be the same for both portions of water, we can set the total heat lost by the hot water equal to the total heat gained by the cold water:

$$Q_{\text{hot}} = Q_{\text{cold}}$$

$$m_{\text{hot}}cΔT_{\text{hot}} = m_{\text{cold}}cΔT_{\text{cold}}$$

Given that the masses of the water portions are the same (both 2 liters), we can simplify this equation to:

$$ΔT_{\text{hot}} = ΔT_{\text{cold}}$$

Now, we can solve for the change in temperature ($ΔT$):

$$75 - T_{\text{final}} = T_{\text{final}} - 25$$

$$75 - 25 = 2T_{\text{final}}$$

$$50 = 2T_{\text{final}}$$

$$T_{\text{final}} = \frac{50}{2}$$

$$T_{\text{final}} = 25°C$$

Therefore, the final temperature of the water mixture will be 25°C.