Ive been absent from class because of the flu so i don't have a clue how to figure these out! Help! due tomorrow at 12!
1).Suppose that 135 g of ethanol at 23◦C is mixed with 255 g of ethanol at 75◦C at constant atmospheric pressure in a thermally insulated vessel. What is the ∆Ssys for the process? The specific heat capacity for ethanol is 2.42 J/g K.
Answer in units of J/K
To find the change in entropy (∆Ssys) for this process, we need to use the formula:
∆Ssys = ∆q / T
where ∆q is the heat transferred and T is the temperature.
First, let's find the heat transferred (∆q). Since the vessel is thermally insulated, there is no heat transfer with the surroundings. Therefore, the heat transferred (∆q) is equal to zero.
Next, we need to find the temperature (T) of the mixture. To do that, we can use the concept of heat capacity:
q = m * c * ∆T
where q is the heat transferred, m is the mass, c is the specific heat capacity, and ∆T is the change in temperature.
For the first container with 135g of ethanol at 23◦C:
q1 = 135g * 2.42 J/g K * (Tf - 23)
For the second container with 255g of ethanol at 75◦C:
q2 = 255g * 2.42 J/g K * (Tf - 75)
Since the two containers exchange heat until they reach thermal equilibrium, q1 must be equal in magnitude but opposite in sign to q2:
|q1| = |q2|
135g * 2.42 J/g K * (Tf - 23) = -(255g * 2.42 J/g K * (Tf - 75))
Now, we can solve this equation for Tf, which is the final temperature of the mixture.
Once we find the final temperature Tf, we can substitute it into the formula for ∆Ssys. As we calculated before, ∆q is equal to zero. Therefore, ∆Ssys will also be zero.
Therefore, the ∆Ssys for this process is 0 J/K.