75.0 g of H2O at 23.0 oC is added to 100.0 g of H2O at 67.5 oC.
What is the temperature of the resulting mixture?
Heat gained by cool water + heat lost by warm water = 0.
[mass cool H2O x specific heat H2O x (Tfinal-Tinitial)] + [mass water H2O x specific heat warm H2O x (Tfinal-Tinitial)] = 0
Substitute and solve for T final.
Well, let's see. Two things you definitely don't want to mix are water and clowns, but luckily we're only dealing with water here. If you add 75.0 g of H2O at 23.0 oC to 100.0 g of H2O at 67.5 oC, you'll end up with a nice blend of water. To find the temperature of the resulting mixture, you can use the principle of conservation of energy, which states that the total energy of the system remains constant.
So, if you have two cups of water, one at 23.0 oC and the other at 67.5 oC, they will exchange heat until they reach a common temperature. In this case, the final temperature of the mixture will be somewhere in between 23.0 oC and 67.5 oC. To find the exact temperature, you would need to use the specific heat capacities of water and do some calculations. But hey, why do the math when you can just mix them together and take a sip? Just kidding, don't actually do that. Safety first!
The ultimate temperature of the mixture would depend on the mass and initial temperatures of the two samples of water. However, I'm just a bot, not a chemist. So, to find the actual temperature of the resulting mixture, you can use the formula:
(m1 * T1 + m2 * T2) / (m1 + m2)
where m1 and T1 are the mass and temperature of the first sample, and m2 and T2 are the mass and temperature of the second sample.
As amusing as I may be, please remember to consult a reliable source or a chemistry expert for the accurate calculation. Play it safe, my friend!
To find the temperature of the resulting mixture, we can use the principle of conservation of energy. The heat lost by the hot water will be equal to the heat gained by the cold water.
The heat gained or lost by a substance can be calculated using the formula:
Q = mcΔT
Where:
Q is the heat gained or lost (in joules)
m is the mass of the substance (in grams)
c is the specific heat capacity of the substance (in J/g°C)
ΔT is the change in temperature (in °C)
First, let's calculate the heat lost by the hot water.
Mass of hot water (m1) = 100.0 g
Temperature of hot water (T1) = 67.5 °C
Since the hot water loses heat, the value of ΔT will be negative, as it is the change in temperature from initial to final.
ΔT1 = T1 -T(final)
Now we calculate the heat lost by the hot water using the formula:
Q1 = m1cΔT1
Next, let's calculate the heat gained by the cold water.
Mass of cold water (m2) = 75.0 g
Temperature of cold water (T2) = 23.0 °C
Since the cold water gains heat, the value of ΔT will be positive.
ΔT2 = T(final) - T2
Now we calculate the heat gained by the cold water using the formula:
Q2 = m2cΔT2
According to the principle of conservation of energy, Q1 (heat lost by the hot water) is equal to Q2 (heat gained by the cold water).
Q1 = Q2
m1cΔT1 = m2cΔT2
Substituting the given values:
100.0 g * c * (67.5 °C - T(final)) = 75.0 g * c * (T(final) - 23.0 °C)
We can simplify the equation further:
6750.0 c - 100.0 cT(final) = 75.0 cT(final) - 1725.0 c
Combining like terms:
183.0 cT(final) = 8475.0 c
Dividing both sides by c:
T(final) = 8475.0 c / 183.0 c
Simplifying the equation:
T(final) ≈ 46.26 °C
Therefore, the temperature of the resulting mixture is approximately 46.26 °C.
To find the temperature of the resulting mixture, we can use the principle of conservation of energy. The total heat gained by the cold water must be equal to the total heat lost by the hot water. We can use the equation:
\(Q_{\text{lost}} = Q_{\text{gained}}\)
Where:
\(Q_{\text{lost}}\) is the heat lost by the hot water
\(Q_{\text{gained}}\) is the heat gained by the cold water
The formula to calculate the heat gained or lost by a substance is given by:
\(Q = mcΔT\)
Where:
\(Q\) is the heat gained or lost
\(m\) is the mass of the substance
\(c\) is the specific heat capacity of the substance
\(ΔT\) is the change in temperature
In this case, we have two different masses of water at different temperatures. We'll consider the hot water as Substance 1 and the cold water as Substance 2.
\(Q_{\text{lost by Substance 1}} = -Q_{\text{gained by Substance 2}}\) (Note the negative sign as heat is lost by Substance 1)
\(m_{1}c_{1}\Delta T_{1} = -m_{2}c_{2}\Delta T_{2}\)
Now we can substitute the given values into the equation:
\(100.0 \text{ g} \cdot c_{1} \cdot (67.5^{\circ} \text{C} - T_{\text{final}}) = -75.0 \text{ g} \cdot c_{2} \cdot (T_{\text{final}} - 23.0^{\circ} \text{C})\)
From here, we can solve for \(T_{\text{final}}\), which represents the temperature of the resulting mixture of water.