5. Assume you have a 45°C 100g block of gold (cp=0.129J/g°C) and a 150°C , 150g block of silver (cp=0.240J/g°C)(1). You place these blocks into an adiabatic container in thermal contact.
a. Calculate the final temperature
b. Calculate the heat transferred
c. Describe how this is consistent with the first law.
d. Would it be consistent with the 1st law if the gold block cooled and the silver block warmed?
6. Assume you have a 3 mole sample of ideal gas at 1.3atm and 45 °C. Calculate q, w, and ∆U if it is expanded:
a. Reversibly and isothermally to 0.8atm
b. Irreversibly and isothermally against an external pressure of 0.7atm until the volume is 98L
7. Taking the gas in question 6 repeat the process of expanding it isothermally against the external pressure of 0.7atm but do this is 4 steps. Calculate q, w, and ∆U. Comment on the different values you attain in 6a, b and 7.
a. Step 1 expand to 70L against an external P of 1.1atm
b. Step 2 expand to 80L against an external P of 0.9atm
c. Step 3 expand to 90L against an external P of 0.8atm
d. Step 4 expand to 98L against an external P of 0.9atm
a.
[mass Au x specific heat Au x (Tfinal-Tinitial)] + [mass Ag x specific heat Ag x (Tfinal-Tintitial)] = 0
Substitute and solve for Tfinal.
b.
q = [mass Au x specific heat Au x (Tfinal-Tinitial)]
OR
q = ([mass Ag x specific heat Ag x (Tfinal-Tinitial)]
q is the same for each the negative of each other.
c. If you plug in the numbers for part b, one will be + and the other will be the negative of the first. Heat lost = heat gained and that is consistent.
d.
Look at the numbers for c. You will know which cooled and which warmed.
To answer questions 5, 6, and 7, we need to use the first law of thermodynamics. The first law states that the change in internal energy (ΔU) of a system is equal to the heat transferred (q) into the system minus the work done (w) by the system.
In order to calculate the final temperature in question 5a, we can use the principle of energy conservation. The total energy in the system will be conserved, so the heat lost by the hot block (gold) will equal the heat gained by the cold block (silver). The equation we can use is:
(q1/gold) + (q2/silver) = 0
where q1 is the heat lost by the gold block, q2 is the heat gained by the silver block. Since the container is adiabatic, there is no transfer of heat to or from the surroundings, so the heat lost or gained by each block is the same as the change in internal energy, given by:
q = m * c * ΔT
where q is the heat (or change in internal energy), m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Using this equation, we can calculate the heat transferred (q) in question 5b. Then we can use that value to calculate the change in internal energy (ΔU = q) for each process described in questions 6 and 7.
Now, let's go through each question step by step:
5a. To calculate the final temperature, we need to equate the heat lost by the hot block to the heat gained by the cold block and solve for the final temperature.
- Calculate the heat lost by the gold block using the equation q = m * c * ΔT.
- Calculate the heat gained by the silver block using the same equation.
- Set the two equations equal to each other and solve for the final temperature.
5b. To calculate the heat transferred, we need to calculate the change in internal energy (ΔU), which is equal to the heat transferred (q). Use the equation q = m * c * ΔT for each block separately, and then sum up the heat transferred by both blocks.
5c. The first law of thermodynamics states that the change in internal energy (ΔU) of a system is equal to the heat transferred (q) into the system minus the work done (w) by the system. In this case, since the container is adiabatic (no heat transfer to or from the surroundings), the heat transferred (q) is equal to the change in internal energy (ΔU). So the first law is consistent in this scenario.
5d. If the gold block cooled and the silver block warmed, the heat transfer between them would have the opposite sign compared to the previous scenario. In this case, the heat transferred would be negative for the gold block and positive for the silver block. Applying the first law of thermodynamics would still give consistent results, but the sign of the heat transferred (q) and the change in internal energy (ΔU) would be reversed.
6a. To calculate q, w, and ΔU for the reversible isothermal expansion, we can first calculate the work done using the equation w = -P * ΔV, where P is the external pressure and ΔV is the change in volume. Since the process is isothermal, the temperature doesn't change, so ΔU is zero. The heat transferred (q) is equal to the absolute value of the work done (q = |w|).
6b. To calculate q, w, and ΔU for the irreversible isothermal expansion, we can use the same equation for work done as in the previous case. The heat transferred (q) is equal to the absolute value of the work done (q = |w|). Since the process is isothermal, the change in internal energy (ΔU) is zero.
7. Since the isothermal expansion is done in four steps, we need to calculate q, w, and ΔU for each step separately. The total q, w, and ΔU will be the sum of the values calculated for each step. The process is still isothermal, so the change in internal energy (ΔU) is zero. The work done (w) can be calculated using the equation w = -P * ΔV, where P is the external pressure and ΔV is the change in volume for each step. The heat transferred (q) is equal to the absolute value of the work done (q = |w|).
Comparing 6a, 6b, and 7, you will notice that the values of q and w might be different in each case, but the change in internal energy (ΔU) is zero for all isothermal processes. This is consistent with the first law of thermodynamics, which states that the change in internal energy (ΔU) is equal to the heat transferred (q) minus the work done (w).