If you have 320.0 mL of water at 25.00 °C and add 110.0 mL of water at 95.00 °C, what is the final temperature of the mixture? Use 1.00 g/mL as the density of water.
Because you're just mixing water, the specific heat is the same for both liquids. So, it's just like adding a bunch of weighted numbers:
(320*25 + 110*95)/(320+110) = 42.91°C
You can use a fancy formula incorporating the specific heats, but they all cancel out in the end, since they're the same.
As a ballpark check, the average of 25 and 95 is 60°C, but you have more cool water than hot, so the result will be below the average.
Steve is right
Here's a little variation of this one:
"If you combine 310.0 mL of water at 25.00 °C and 100.0 mL of water at 95.00 °C, what is the final temperature of the mixture? Use 1.00 g/mL as the density of water."
ANS: 42.07 Degrees
If you combine 310.0 mL of water at 25.00 °C and 140.0 mL of water at 95.00 °C, what is the final temperature of the mixture? Use 1.00 g/mL as the density of water.
Well, mixing water at different temperatures can be quite the chilling experience, don't you think? Let's dive into this puzzle.
To find the final temperature, we can use the conservation of energy principle. The total heat gained by the cold water should equal the total heat lost by the hot water.
Now, to calculate the heat gained or lost by a substance, we can use the formula:
Q = m * c * ΔT
Where:
Q = heat gained/lost (in joules)
m = mass (in grams)
c = specific heat capacity (in J/g°C)
ΔT = change in temperature (in °C)
First, let's calculate the heat lost by the hot water (Q_hot):
Q_hot = m_hot * c * ΔT_hot
m_hot = density * volume_hot
density = 1.00 g/mL
volume_hot = 110.0 mL
ΔT_hot = T_final - 95.00 °C
Now, let's calculate the heat gained by the cold water (Q_cold):
Q_cold = m_cold * c * ΔT_cold
m_cold = density * volume_cold
density = 1.00 g/mL
volume_cold = 320.0 mL
ΔT_cold = T_final - 25.00 °C
Since the heat gained by the cold water equals the heat lost by the hot water, we have:
Q_hot = Q_cold
m_hot * c * ΔT_hot = m_cold * c * ΔT_cold
Now, let's substitute the expressions we derived earlier:
(density * volume_hot) * c * (T_final - 95.00 °C) = (density * volume_cold) * c * (T_final - 25.00 °C)
Now, we can cancel out the density and specific heat capacity:
volume_hot * (T_final - 95.00°C) = volume_cold * (T_final - 25.00°C)
Plug in the values:
110.0 mL * (T_final - 95.00°C) = 320.0 mL * (T_final - 25.00°C)
Now, let's do some algebraic acrobatics to solve for T_final:
110.0 * T_final - 110.0 * 95.00 = 320.0 * T_final - 320.0 * 25.00
110.0 * T_final - 11000.00 = 320.0 * T_final - 8000.00
-11000.00 + 8000.00 = 320.0 * T_final - 110.0 * T_final
-3000.00 = 210.0 * T_final
T_final = -3000.00 / 210.0
T_final ≈ -14.29 °C
Oh, dear! It seems like the final temperature I calculated is negative. I apologize for the chilly surprise. There might have been an error in the calculations. Let's give it another shot, shall we?
To find the final temperature of the mixture, we can use the principle of conservation of energy, which states that the total energy of a closed system remains constant.
We can assume that no heat is lost to the surroundings, so the heat lost by the hot water will be equal to the heat gained by the cold water.
The equation we will use is:
Q_lost = Q_gained
where Q_lost is the heat lost by the hot water, and Q_gained is the heat gained by the cold water.
The formula to calculate the heat lost or gained is:
Q = m * c * ΔT
where Q is the heat gained or lost, m is the mass of the water, c is the specific heat capacity of water (which is approximately 4.184 J/g°C), and ΔT is the change in temperature.
Let's start by calculating the mass of the hot water:
mass_hot_water = density * volume = 1.00 g/mL * 110.0 mL = 110.0 g
Now we can calculate the heat lost by the hot water:
Q_lost = m * c * ΔT = 110.0 g * 4.184 J/g°C * (95.00 °C - final_temperature)
Next, let's calculate the mass of the cold water:
mass_cold_water = density * volume = 1.00 g/mL * 320.0 mL = 320.0 g
Now we can calculate the heat gained by the cold water:
Q_gained = m * c * ΔT = 320.0 g * 4.184 J/g°C * (final_temperature - 25.00 °C)
Using the principle of conservation of energy, we can equate the two equations:
110.0 g * 4.184 J/g°C * (95.00 °C - final_temperature) = 320.0 g * 4.184 J/g°C * (final_temperature - 25.00 °C)
Now, we can solve this equation to find the final temperature.
110.0 g * (95.00 °C - final_temperature) = 320.0 g * (final_temperature - 25.00 °C)
Simplifying the equation, we get:
10340.0 g°C - 110.0 g * final_temperature = 13312.0 g°C - 320.0 g * final_temperature
Rearranging the equation:
320.0 g * final_temperature - 110.0 g * final_temperature = 13312.0 g°C - 10340.0 g°C
210.0 g * final_temperature = 2972.0 g°C
final_temperature = 2972.0 g°C / 210.0 g
final_temperature = 14.152 g°C
Therefore, the final temperature of the mixture is 14.152 °C.