25.0 mL of ethanol (density = 0.789 g/mL) initially at 7.0 degrees celsius is mixed with 35.0 mL of water (density = 1.0 g/mL) initially at 25.3 degrees celsius in an insulated beaker. Assuming that no heat is lost, what is the final temperature of the mixture?
Use density to convert mL ethanol to grams.
Use density to convert mL H2O to grams.
heat gained by ethanol + heat lost by H2O = 0
[mass ethanol x specific heat ethanol x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Substitute and solve for Tfinal.
Well, well, well! Looks like we have a little chemistry problem here. Mixing ethanol and water, eh?
To solve this temperature puzzle, we can use the principle of conservation of energy. Since the beaker is insulated and no heat is lost, we can assume that the heat gained by the water is equal to the heat lost by the ethanol.
First, let's calculate the heat gained by the water. The specific heat capacity of water is about 4.18 J/g°C. So, the heat gained by the water can be calculated using the formula:
Q = m x c x ΔT
Where:
Q is the heat gained
m is the mass of water
c is the specific heat capacity of water
ΔT is the change in temperature
Now, let's find the mass of water:
mass = density x volume
mass = 1.0 g/mL x 35.0 mL
Once you calculate the mass of the water and plug in the values, you can solve for Q.
Next, we can calculate the heat lost by the ethanol using the same formula. The specific heat capacity of ethanol is about 2.44 J/g°C.
Use the density of ethanol and the volume provided to find the mass of ethanol, and then calculate the heat lost by the ethanol using the formula above.
Now, since we have assumed no heat is lost, we can equate the heat gained by the water to the heat lost by the ethanol:
Q_water = Q_ethanol
Once you have that equation, you can solve for the final temperature (ΔT) of the mixture.
Please give it a try, and let me know if you need any further assistance!
To find the final temperature of the mixture, we can use the principle of heat transfer, specifically the equation:
Q = mcΔT
where Q is the heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
To solve this problem, we need to apply this equation separately to ethanol and water, and then combine the results.
First, let's calculate the heat transferred for ethanol and water separately using the equation above.
For ethanol:
- Mass (m) = Volume (V) * Density (ρ) = 25.0 mL * 0.789 g/mL = 19.725 g
- Specific heat capacity (c) of ethanol is approximately 2.44 J/g°C (rounded)
- Initial temperature (T1) = 7.0°C
- Final temperature (T2) is what we want to find
For water:
- Mass (m) = Volume (V) * Density (ρ) = 35.0 mL * 1.0 g/mL = 35.0 g
- Specific heat capacity (c) of water is approximately 4.18 J/g°C (rounded)
- Initial temperature (T1) = 25.3°C
- Final temperature (T2) is what we want to find
Now, we can apply the equation to calculate the heat transferred for each substance:
Q_ethanol = m_ethanol * c_ethanol * (T2 - T1)
Q_water = m_water * c_water * (T2 - T1)
Since the beaker is insulated, the heat lost by the ethanol will equal the heat gained by the water. Therefore, Q_ethanol = -Q_water.
Substituting the values, we have:
m_ethanol * c_ethanol * (T2 - T1) = -m_water * c_water * (T2 - T1)
Rearranging the equation, we get:
[T2 - T1] * [m_ethanol * c_ethanol + m_water * c_water] = 0
Dividing both sides by [m_ethanol * c_ethanol + m_water * c_water], we can solve for T2:
T2 = T1
So, the final temperature of the mixture is equal to the initial temperature of the ethanol and water, which is 7.0 degrees Celsius.