Final answer:
The problem involves finding the volume of ethanol at 87.7°C that needs to be mixed with 238 mL of ethanol at 55.5°C to reach a final temperature of 76.3°C. By applying the conservation of energy principle and assuming no heat loss to the surroundings, it is calculated that approximately 438 mL of ethanol is needed.
Step-by-step explanation:
To find the volume of ethanol that needs to be mixed to achieve a final temperature of 76.3°C, we can assume that the heat lost by the warmer ethanol is gained by the cooler ethanol and no heat is lost or gained from the surroundings.
This is an application of the principle of conservation of energy where the heat (q) exchanged is equal to the mass (m), the specific heat capacity of ethanol (c, which we assume to be 2.44 J/g°C for liquid ethanol), and the change in temperature (ΔT). Since both substances are ethanol, their specific heats are the same and cancel out, allowing us to use the simpler equation q = mcΔT for calculations.
Let V be the volume of ethanol initially at 87.7°C. Because ethanol has a density of approximately 0.789 g/mL, we can convert volumes to mass (m). Using the given temperatures, we have:
- Heat lost by warmer ethanol: q = (0.789 g/mL * V) * 2.44 J/g°C * (87.7°C - 76.3°C)
- Heat gained by cooler ethanol: q = (0.789 g/mL * 238 mL) * 2.44 J/g°C * (76.3°C - 55.5°C)
Equating the heat lost to heat gained and simplifying, we solve for V:
(0.789 g/mL * V) * 2.44 J/g°C * (11.4°C) = (0.789 g/mL * 238 mL) * 2.44 J/g°C * (20.8°C)
After solving, we find that the volume of ethanol required is:
V = ≈ 438 mL