When 4.62 mL of cold water at 22.6 °C is mixed with 37.5 mL of hot water at 67.9 °C in a thermally isolated calorimeter, what is the final temperature of the mixture when thermal equilibrium has been achieved? H2O density = 1.00 g/mL.
heat lost by warm water + heat gained by cool water = 0
[mass warm water x specific heat water x (Tfinal-Tinitial)] + [mass cool water x specific heat water x (Tfinal-Tinitital)] = 0
Substitute and solve for Tf.
Note that with density H2O of 1.00 g/mL that makes mL H2O the same as grams H2O.
[37.5*4.18*(Tf-67.9)+4.62*4.18*(Tf-22.6)
156.75(Tf-67.9)+19.3116(Tf-22.6)
(156.75Tf-10643.325)+(19.3116Tf-436.44216)
176.0616Tf=11079.76716
Tf=11079.76716/176.0616
Tf=62.9
62.9°C
To find the final temperature of the mixture when thermal equilibrium has been achieved, you can use the principle of conservation of energy, which states that energy gained by one object is equal to the energy lost by another object. In this case, the energy gained by the cold water will be equal to the energy lost by the hot water.
The equation you can use is:
(mass of cold water) × (specific heat capacity of cold water) × (change in temperature of cold water) = (mass of hot water) × (specific heat capacity of hot water) × (change in temperature of hot water)
Let's break down the information given and solve the equation step by step.
1. Convert the volumes of water to masses using the density of water. The density of water is given as 1.00 g/mL. Therefore:
- The mass of cold water = 4.62 mL × 1.00 g/mL = 4.62 g
- The mass of hot water = 37.5 mL × 1.00 g/mL = 37.5 g
2. Determine the specific heat capacities of cold water and hot water. The specific heat capacity of water is approximately 4.18 J/g°C.
3. Calculate the change in temperature for the cold water and hot water by subtracting their initial temperatures from the final temperature, assuming the final temperature is denoted as "Tf":
- Change in temperature of cold water = Tf - 22.6°C
- Change in temperature of hot water = Tf - 67.9°C
4. Substitute the known values into the equation:
(4.62 g) × (4.18 J/g°C) × (Tf - 22.6°C) = (37.5 g) × (4.18 J/g°C) × (Tf - 67.9°C)
5. Simplify the equation:
19.2636 Tf - 92.3412 = 156.135 Tf - 5310.15
6. Rearrange the equation to solve for Tf:
136.8724 Tf = 5217.8088
7. Divide both sides by 136.8724 to isolate Tf:
Tf ≈ 38.10°C
Therefore, the final temperature of the mixture, when thermal equilibrium is achieved, is approximately 38.10°C.