New problem. Any assistance is greatly appreciated.
22.5 ml of ethanol (density=0.789 g/ml) initially at 7.7 deg. C is mixed with 31.6 ml of water (density=1.0 g/ml) initially at 27.1 deg. C in an insulated beaker. Assuming that no heat is lost, what is the final temperature of the mixture. Express your answer using two significant figures.
22.5ml EtOH 22.5ml x 0.789g/1ml= 17.75 g
31.6ml water 31.6ml x 1g/ml=31.6g
total mass= 49.4 g
This is where I am at a loss. There is no q given. q= mass x SHC x delta temp.
SHC ethanol 2.42 J/g*C
SHC water 4.184 J/g*C
Maybe I am making this more complicated than it should be. Any direction now is appreciated. Since there is no heat lost, do I assume that q=1 or do I just subtract the two temps. 27.1*C - 7.7*C=
To find the final temperature of the mixture, you need to consider the heat transferred between the ethanol and water.
As you correctly mentioned, the heat transferred (q) can be calculated using the formula: q = mass x specific heat capacity (SHC) x change in temperature (ΔT).
Since there is no heat lost to the surroundings, the amount of heat gained by the water equals the amount of heat lost by the ethanol. Therefore, q1 = -q2.
To solve the problem, you can follow these steps:
1. Calculate the heat gained by the water (q1):
q1 = mass_water x SHC_water x ΔT_water
mass_water = volume_water x density_water
ΔT_water = final temperature - initial temperature of water
2. Calculate the heat lost by the ethanol (q2):
q2 = mass_ethanol x SHC_ethanol x ΔT_ethanol
mass_ethanol = volume_ethanol x density_ethanol
ΔT_ethanol = final temperature - initial temperature of ethanol
3. Set up an equation using q1 = -q2 and solve for the final temperature.
Now, let's calculate it step by step:
Step 1:
mass_water = 31.6 ml x 1.0 g/ml = 31.6 g
ΔT_water = final temperature - initial temperature of water = T - 27.1°C
Step 2:
mass_ethanol = 22.5 ml x 0.789 g/ml = 17.77 g
ΔT_ethanol = final temperature - initial temperature of ethanol = T - 7.7°C
Step 3:
q1 = q2
mass_water x SHC_water x ΔT_water = -mass_ethanol x SHC_ethanol x ΔT_ethanol
Substituting the values:
31.6 g x 4.184 J/g°C x (T - 27.1°C) = -17.77 g x 2.42 J/g°C x (T - 7.7°C)
Simplifying the equation:
131.9744 x (T - 27.1) = -43.0534 x (T - 7.7)
Solving for T:
131.9744T - 3565.7424 = -43.0534T + 332.4622
175.0278T = 3898.2046
T ≈ 22.27°C
Therefore, the final temperature of the mixture is approximately 22.27°C.