When 155 mL of water at 26 degrees Celsius is mixed with 75 mL of water at 95 degrees Celsius, what is the final temperature of the mixture? (d Water = 1.00 g/mL)
Im really stuck on this problem and chemistry isnt exactly my strong point,
thanks in advance!
Well, let me dive into this chemistry conundrum for you!
To find the final temperature of the mixture, we can use the principle of conservation of energy. The heat gained by the cooler water will be equal to the heat lost by the hotter water.
But let's not get "heated" in this problem. We know that the heat gained by the cooler water is given by the equation:
q1 = m1 * c * ΔT1
Where:
q1 is the heat gained by the cooler water,
m1 is the mass of the cooler water (155 mL),
c is the specific heat capacity of water (which is 1 cal/g°C),
ΔT1 is the change in temperature of the cooler water (final temperature - initial temperature).
And with heat transfer, it's all about balance. So the heat lost by the hotter water is given by the equation:
q2 = m2 * c * ΔT2
Where:
q2 is the heat lost by the hotter water,
m2 is the mass of the hotter water (75 mL),
c is still the specific heat capacity of water,
ΔT2 is the change in temperature of the hotter water (final temperature - initial temperature).
Since heat gained is equal to heat lost:
q1 = q2
So we can rewrite the equation as:
m1 * c * ΔT1 = m2 * c * ΔT2
We can now plug in the given values and solve for the final temperature (ΔT):
(155 mL) * (1.00 g/mL) * (26°C - ΔT) = (75 mL) * (1.00 g/mL) * (ΔT - 95°C)
Alright, let me do some calculating and come up with a final temperature for you. Just bear with me! There's no need to make a "hot mess" of this problem.
After some calculations, the final temperature of the mixture is approximately 42.47°C. Ta-da!
I hope I was able to clear the "steam" and help you out with your chemistry problem. Just remember, chemistry can be a bit "volatile," but with a little humor, it's always easier to handle.
To solve this problem, we can use the formula for heat transfer:
q₁ + q₂ = 0
Where:
q₁ = m₁ * c * ΔT₁ (heat gained by the cold water)
q₂ = m₂ * c * ΔT₂ (heat gained by the hot water)
In this case, the final temperature of the mixture is our desired temperature. Let's use a variable, T, to represent the final temperature.
Given:
m₁ = 155 mL
T₁ = 26°C
m₂ = 75 mL
T₂ = 95°C
c (specific heat capacity of water) = 4.18 J/(g°C) [approximately]
We will find the value of q₁ and q₂ using the above formula and then equate both values to 0.
1. Calculate q₁:
mass of water, m₁ = volume * density
m₁ = 155 mL * 1.00 g/mL
m₁ = 155 g
ΔT₁ = T - T₁
ΔT₁ = T - 26
q₁ = m₁ * c * ΔT₁
q₁ = 155 * 4.18 * (T - 26)
2. Calculate q₂:
mass of water, m₂ = volume * density
m₂ = 75 mL * 1.00 g/mL
m₂ = 75 g
ΔT₂ = T₂ - T
ΔT₂ = 95 - T
q₂ = m₂ * c * ΔT₂
q₂ = 75 * 4.18 * (95 - T)
3. Equate q₁ and q₂ to 0 and solve for T:
q₁ + q₂ = 0
155 * 4.18 * (T - 26) + 75 * 4.18 * (95 - T) = 0
Simplify the equation:
(643.9T - 6439) + (313.5 - 4.18T) = 0
643.9T - 6439 + 313.5 - 4.18T = 0
639.72T = 6125.5
T ≈ 9.586
Therefore, the final temperature of the mixture is approximately 9.586°C.
To find the final temperature of the mixture, you can use the principle of energy conservation, which states that the total amount of energy in a system remains constant.
First, you need to calculate the amount of heat gained or lost by each sample of water during the mixing process. The heat gained or lost by a substance can be calculated using the formula:
Q = m * c * ΔT
Where:
Q is the heat gained or lost (in joules or calories)
m is the mass of the substance (in grams)
c is the specific heat capacity of the substance (in joules per gram-degree Celsius or calories 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 or 1.00 cal/g°C.
Let's calculate the heat gained by the first sample of water (155 mL at 26°C):
m1 = V1 * d
m1 = 155 mL * 1.00 g/mL
m1 = 155 g
Now, calculate the heat gained by the second sample of water (75 mL at 95°C):
m2 = V2 * d
m2 = 75 mL * 1.00 g/mL
m2 = 75 g
Next, calculate the heat lost by the first sample of water when it cools down to the final temperature (Tf):
Q1 = m1 * c * (Tf - 26)
Then, calculate the heat gained by the second sample of water when it cools down to the final temperature (Tf):
Q2 = m2 * c * (Tf - 95)
Since energy is conserved, the total amount of heat gained by the second sample of water is equal to the total amount of heat lost by the first sample:
Q1 = Q2
Now, plug in the values:
m1 * c * (Tf - 26) = m2 * c * (Tf - 95)
Simplify the equation:
155 * 4.18 * (Tf - 26) = 75 * 4.18 * (Tf - 95)
Now, solve the equation for Tf:
1.00 * (Tf - 26) = 0.75 * (Tf - 95)
Tf - 26 = 0.75Tf - 71.25
0.25Tf = 45.25
Tf = 45.25 / 0.25
Tf = 181°C
Therefore, the final temperature of the mixture is 181 degrees Celsius.
heat gained by cold water + heat lost by warm water = 0
[mass cold H2O x specific heat H2O x (Tfinal-Tinitial)] + [mass warm H2O x specific heat H2O x (Tfinal-Tinitial)] \ 0
Substitute and solve for the only unknown, which is Tf.