An unknown volume of water at 18.2 degrees celsius is added to 24.4 mL of water at 35.0 degrees celsius. If the final temperature is 23.5 degrees celsius what was the unknown volume? (assume no heat is lost to surroundings and D of water is 1.00g/mL)
V = unknown volume of H2O, which equals the mass because the density is 1.00
V(4.184)(23.5-18.2) + 24.4(4.184)(23.5-35) = 0
22.1752V = 1174.0304
V = 52.94mL
Heat gained by water + heat lost by water = 0.
unknown mass H2O x specific heat x (Tf-Ti) + known mass water x specific heat x (Tf-Ti) = 0
Tf = final T
Ti = initial T.
You have only one unknown; i.e., unknown mass H2O. Post your work if you get stuck.
Oh, it's getting steamy in here! So we've got two temperatures and a final temperature, and we want to find the unknown volume of water. Let's dive right in!
First, let's calculate the heat gained or lost by the known volume of water using the formula:
Q = m * c * ΔT
where Q is the heat gained or lost, m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature.
Since the density of water is 1.00g/mL, the mass of the known volume of water is 24.4g (24.4 mL * 1.00g/mL). The specific heat capacity of water is approximately 4.18 J/g°C.
Using the initial and final temperatures, we can calculate the heat gained or lost by the known volume of water. Let's do some math!
Q1 = 24.4g * 4.18 J/g°C * (23.5°C - 35.0°C)
Now, to find the unknown volume of water, we can use the equation:
Q1 = Q2
where Q1 is the heat gained or lost by the known volume of water and Q2 is the heat gained or lost by the unknown volume of water.
Since no heat is lost to the surroundings, we can set Q1 equal to Q2.
24.4g * 4.18 J/g°C * (23.5°C - 35.0°C) = m * c * (23.5°C - 18.2°C)
Now, we can solve for the unknown volume, represented by "m". Let's grab our calculators and find that missing volume!
To solve this problem, we can use the principle of heat transfer known as the heat equation:
heat gained by cold substance = heat lost by hot substance
The heat gained by the cold water can be calculated by the equation:
Qcold = mcΔT
Where:
Qcold is the heat gained by the cold water (unknown volume),
m is the mass of the cold water,
c is the specific heat capacity of water (which is approximately 4.18 J/g·°C),
ΔT is the change in temperature of the cold water.
The heat lost by the hot water can be calculated using the same equation:
Qhot = mcΔT
Where:
Qhot is the heat lost by the hot water (24.4 mL),
m is the mass of the hot water,
c is the specific heat capacity of water (which is approximately 4.18 J/g·°C),
ΔT is the change in temperature of the hot water.
Since no heat is lost to the surroundings, the heat gained by the cold water must be equal to the heat lost by the hot water.
So, we can set up an equation:
mcoldΔTcold = mhotΔThot
Now, we'll plug in the values we know:
mhot = 24.4 mL * 1.00 g/mL (density of water is 1.00 g/mL) = 24.4 g
ΔThot = 35.0°C - 23.5°C = 11.5°C
mcold = unknown volume * 1.00 g/mL (density of water is 1.00 g/mL)
ΔTcold = 23.5°C - 18.2°C = 5.3°C
After substituting these values into the equation, we can solve for the unknown volume:
(unknown volume * 1.00 g/mL) * 5.3°C = 24.4 g * 11.5°C
To isolate the unknown volume, we'll divide both sides by 5.3 g/°C:
unknown volume * 1.00 g/mL = 24.4 g * 11.5°C / 5.3 g/°C
Now, we'll solve for the unknown volume:
unknown volume = (24.4 g * 11.5°C / 5.3 g/°C) / (1.00 g/mL)
Calculating this expression will give you the unknown volume of water that was added.
x= unknown mass of H2O
x(4.184)(23.5-35) + 17.008(4.184)(23.5-18.2)=0
-48.116x + 377.156= 0
x = 7.84g
d=1.00g/mL x=unknown volume
1.00= 7.84/x
x=7.84mL
x= unknown mass of H2O
x(4.184)(23.5-35) + 17.008(4.184)(23.5-18.2)=0
I don't think so. Tf = final T; Ti = initial T.
a. Isn't the initial T 18.2 and not 35 for the unknown amount of water?
b. Where did the 17.008 come from? That must be the mass of water but that was 24.4 mL so it has a mass of 24.4 mL x 1.00 g/mL = 24.4 grams.
c. And delta T for the known amount of water is 23.5 - 35.0 isn't it?
-48.116x + 377.156= 0
x = 7.84g
d=1.00g/mL x=unknown volume
1.00= 7.84/x
x=7.84mL