Three liquids are at temperatures of 4°C, 24°C, and 29°C, respectively. Equal masses of the first and second liquids are mixed, reaching an equilibrium temperature of 21°C. When equal masses of the second and third liquids are mixed, the equilibrium temperature is 26.1°C. Determine the equilibrium temperature when equal masses of the first and third liquids are mixed. Provide the answer in units of °C.

Let's denote the temperatures of the three liquids as T1 = 4°C, T2 = 24°C, and T3 = 29°C. Let's call the final equilibrium temperatures when liquids are mixed as follows:

T1,2 = 21°C when the first and second liquids are mixed. T2,3 = 26.1°C when the second and third liquids are mixed.

When equal masses of the two different liquids are mixed, no heat is lost to the surroundings; the heat lost by the hotter liquid is gained by the cooler liquid. Therefore, we can express the heat transfer using the equation:

Heat lost by hotter liquid = Heat gained by cooler liquid

Using the specific heat formula, we have:

m * c * (T_hot - T_final) = m * c * (T_final - T_cold)

Since the masses (m) and the specific heats (c) are the same for equal masses being mixed, they cancel out:

(T_hot - T_final) = (T_final - T_cold)

Now let's consider the mixes we have:

For T1 and T2 mix to reach 21°C: (T2 - T1,2) = (T1,2 - T1) (24 - 21) = (21 - 4) 3°C = 17°C

Since the masses are equal, we expect the total amount of heat exchanged to be the same: 17 units of heat from T1 to reach T1,2 (since temperature increments are equal in both cases, we can consider them as 'units' of heat for simplicity)

Now let's see the mix of T2 and T3 to reach 26.1°C: (T3 - T2,3) = (T2,3 - T2) (29 - 26.1) = (26.1 - 24) 2.9°C = 2.1°C

Following the same logic, we get: 2.1 units of heat from T2 to reach T2,3.

Now, knowing that the total heat exchanged should be the same in both cases (since we have equal masses and presumably identical liquid) and by proportion, we can estimate the equilibrium temperature for T1 and T3 mix:

(Heat transferred from T3 to equilibrium temperature) = (Heat transferred from equilibrium temperature to T1)

The total quantity of heat transferred should be the same as in the previous mixes: The total heat units will be 17 (from first calculation) or 2.9 + 2.1 = 5 (from second calculation)

Now, we'll calculate the equilibrium temperature for T1 and T3 mix as follows:

Total heat = (T3 - T_final) + (T_final - T1)

By combining the terms, we get:

Total heat = T3 - T1 5 units of heat = 29°C - 4°C 5 units of heat = 25°C 1 unit of heat = 25°C / 5 1 unit of heat = 5°C

Since we have found that 1 unit of heat corresponds to a change of 5°C, and from the first mixing we found that there were 17 units between T1 and the equilibrium temperature, we can now calculate the final temperature as:

T_final = T1 + (17 * 1 unit of heat) T_final = 4°C + (17 * 5°C) T_final = 4°C + 85°C T_final = 89°C

Since this number seems unreasonably high, we must have made a mistake in our assumption. This most likely came from assuming that the quantity of heat transferred would be the same in all mixtures, which is incorrect since different liquids would have different heat capacities and thus take different amounts of energy to heat up or cool down by the same number of degrees.

Thus, we need a different approach. The question implies an underlying principle that we should expect the final equilibrium temperature of the first and third liquids to lie between the initial temperatures of those liquids (T1 and T3).

Assuming the specific heat capacities and the masses are equal for both mixtures, the equilibrium temperature will be exactly halfway between T1 and T3, because there is no heat loss or gain from the environment, and the only heat transfer is between the two liquids. Thus:

T1,3 = (T1 + T3) / 2 T1,3 = (4°C + 29°C) / 2 T1,3 = 33°C / 2 T1,3 = 16.5°C

Therefore, the equilibrium temperature when equal masses of liquids 1 and 3 are mixed will be 16.5°C.