What is the final temperature when 180.0 mL water at 101.3 C is added to 93.1 mL of water at 65.5 C?
To find the final temperature when two substances are mixed, we can use the principle of conservation of energy. The equation we can use is:
(m1 * c1 * ΔT1) + (m2 * c2 * ΔT2) = 0
where:
m1 and m2 are the masses of the substances
c1 and c2 are the specific heat capacities of the substances
ΔT1 and ΔT2 are the temperature changes of the substances
In this case, the substances are both water, so their specific heat capacities are the same (c1 = c2 = 4.18 J/g°C). We also know the initial and final temperatures for each substance. Let's proceed with the calculation step by step.
Step 1: Convert the volumes of water to masses using their density (assumed to be 1 g/mL for water).
- For the first sample:
m1 = 180.0 mL * 1 g/mL = 180.0 g
- For the second sample:
m2 = 93.1 mL * 1 g/mL = 93.1 g
Step 2: Calculate the temperature changes ΔT1 and ΔT2.
- For the first sample:
ΔT1 = final temperature - initial temperature = final temperature - 101.3°C
- For the second sample:
ΔT2 = final temperature - initial temperature = final temperature - 65.5°C
Step 3: Plug the values into the equation and solve for the final temperature.
(180.0 g * 4.18 J/g°C * (final temperature - 101.3°C)) + (93.1 g * 4.18 J/g°C * (final temperature - 65.5°C)) = 0
Simplifying:
753.24 * (final temperature - 101.3) + 388.858 * (final temperature - 65.5) = 0
Combining like terms:
753.24 * final temperature - 753.24 * 101.3 + 388.858 * final temperature - 388.858 * 65.5 = 0
Simplifying further:
1142.098 * final temperature - 76227.492 + 202200.27 - 25511.849 = 0
Combining like terms again:
1497.158 * final temperature = 135538.619
Dividing both sides by 1497.158:
final temperature = 135538.619 / 1497.158
final temperature ≈ 90.5°C
Therefore, the final temperature of the mixture is approximately 90.5°C.
To find the final temperature, we can use the principle of conservation of energy and the equation for heat transfer.
First, let's determine the amount of heat transferred by each sample of water. We can use the equation:
Q = mcΔT
Where:
Q is the heat transferred in joules
m is the mass of the substance in grams
c is the specific heat capacity of the substance in joules per gram degree Celsius
ΔT is the change in temperature in degrees Celsius
For water, the specific heat capacity (c) is approximately 4.18 J/g°C.
Calculating the heat transferred for the first sample of water:
mass = volume × density
density = 1.00 g/mL for water
mass1 = 180.0 mL × 1.00 g/mL = 180.0 g
ΔT1 = final temperature - initial temperature = Tf - 101.3°C
Using the formula Q = mcΔT, we get:
Q1 = (180.0 g) × (4.18 J/g°C) × (Tf - 101.3°C)
Similarly, calculating the heat transferred for the second sample of water:
mass2 = 93.1 mL × 1.00 g/mL = 93.1 g
ΔT2 = final temperature - initial temperature = Tf - 65.5°C
Using the formula Q = mcΔT, we get:
Q2 = (93.1 g) × (4.18 J/g°C) × (Tf - 65.5°C)
Now, since energy is conserved, the total heat transferred by both samples will be equal to zero when they reach the same final temperature. Thus, Q1 + Q2 = 0.
Substituting the earlier equations:
(180.0 g) × (4.18 J/g°C) × (Tf - 101.3°C) + (93.1 g) × (4.18 J/g°C) × (Tf - 65.5°C) = 0
Now we can solve this equation to find the final temperature (Tf).
I'll now calculate the final temperature for you.
heat lost by warm water + heat gained by cold water = 0
[mass warm water x specific heat H2O x (Tfinal-Tinitial)] + [mass cool water x specific heat H2O x (Tfinal-Tinitial)] = 0
Substitute and solve for Tfial