if you have 380.0mL of water at 25.00 degrees Celsius and add 140mL of water at 95.00 degrees Celsius, what is the final temperature of the mixture? Use 1.00 g/mol as the density of water.
heat gained by cool water + heat lost by warm water = 0
[mass cool H2O x specific heat x (Tfinal-Tintial)] + [mass warm water x specific heat x (Tfinal-Tinitial)] = 0
Solve for Tf.
How much heat is required to convert 72.4 of ethanol at -136 to the vapor phase at 78 ?
To calculate the final temperature of the mixture, we can use the principle of conservation of energy and the formula for specific heat capacity.
First, let's calculate the heat gained by the water at 25.00°C (the initial temperature) when it is heated up to the final temperature.
The formula for heat transfer is given by Q = m * c * ΔT, where:
Q is the heat transferred,
m is the mass of the substance,
c is the specific heat capacity of the substance, and
ΔT is the change in temperature.
Let's calculate the heat gained by the water at 25.00°C:
Q1 = m1 * c * ΔT1
where m1 = mass of water at 25.00°C and ΔT1 = final temperature - initial temperature
The mass of the water at 25.00°C can be calculated using the formula:
mass = density * volume
Given:
density of water = 1.00 g/mL
volume = 380.0 mL
mass1 = density * volume
= 1.00 g/mL * 380.0 mL
Substituting the values and solving:
mass1 = 380.0 g
Now, let's calculate the heat gained by the water at 95.00°C:
Q2 = m2 * c * ΔT2
where m2 = mass of water at 95.00°C and ΔT2 = final temperature - initial temperature
The mass of the water at 95.00°C can be calculated using the same formula:
mass2 = density * volume
= 1.00 g/mL * 140 mL
Substituting the values and solving:
mass2 = 140.0 g
Now, in order to find the final temperature of the mixture, we can use the principle of conservation of energy:
Q1 + Q2 = 0
Since heat gained is equal to the heat lost:
Q1 = -Q2
Substituting the values:
m1 * c * ΔT1 = -m2 * c * ΔT2
We know that c is the same for both water samples, so we can cancel it out:
m1 * ΔT1 = -m2 * ΔT2
Substituting the values:
380.0 g * ΔT1 = -140.0 g * ΔT2
Now, rearrange the equation to solve for the final temperature:
ΔT1 = ( -140.0 g * ΔT2 ) / 380.0 g
ΔT1 / ΔT2 = -140.0 g / 380.0 g
ΔT1 / ΔT2 = -0.3684
Now, to find the final temperature, we can use the equation:
final temperature = initial temperature + ΔT1
Substituting the values:
final temperature = 25.00°C + (-0.3684 * ΔT2)
The final temperature of the mixture depends on the value of ΔT2. In order to calculate the specific value, we need more information about the heat capacity of water or the specific value of ΔT2.
To determine the final temperature of the water mixture, we can use the principle of conservation of energy. The total heat gained by the colder water is equal to the total heat lost by the hotter water.
First, let's calculate the initial heat content (Q1) of the colder water:
Q1 = mass * specific heat * temperature change
Since the density of water is given as 1.00 g/mL, the mass of the colder water is:
mass1 = volume1 * density = 380.0 mL * 1.00 g/mL = 380.0 g
The specific heat capacity of water is approximately 4.18 J/g°C.
The temperature change of the colder water is:
temperature change1 = final temperature - initial temperature = final temperature - 25.00°C
So, the initial heat content of the colder water is:
Q1 = 380.0 g * 4.18 J/g°C * (final temperature - 25.00°C)
Next, let's calculate the heat content (Q2) that the hotter water loses:
Q2 = mass2 * specific heat * temperature change
The mass of the hotter water is:
mass2 = volume2 * density = 140 mL * 1.00 g/mL = 140 g
The temperature change of the hotter water is:
temperature change2 = final temperature - initial temperature = final temperature - 95.00°C
Therefore, the heat content lost by the hotter water is:
Q2 = 140 g * 4.18 J/g°C * (final temperature - 95.00°C)
According to the principle of conservation of energy, the total heat gained by the colder water (Q1) must equal the total heat lost by the hotter water (Q2). Therefore, we can equate the equations:
Q1 = Q2
380.0 g * 4.18 J/g°C * (final temperature - 25.00°C) = 140 g * 4.18 J/g°C * (final temperature - 95.00°C)
We can now solve this equation to find the final temperature of the mixture. Let's start by simplifying the equation:
1591.6 g°C * (final temperature - 25.00°C) = 584.8 g°C * (final temperature - 95.00°C)
Now, expand and rearrange the equation:
1591.6 g°C * final temperature - 39,790 g°C = 584.8 g°C * final temperature - 55,516 g°C
Combine like terms:
1591.6 g°C * final temperature - 584.8 g°C * final temperature = -39,790 g°C + 55,516 g°C
1006.8 g°C * final temperature = 15,726 g°C
Finally, solve for the final temperature:
final temperature = 15,726 g°C / 1006.8 g°C
Final temperature = 15.6°C
Therefore, the final temperature of the water mixture is approximately 15.6 degrees Celsius.