Any help would be appreciated!
Which of the following is a way to improve the efficiency of a heat engine?
increase Qh
reduce Qh
reduce Wnet
increase Qc
A
What are the energies associated with atomic motion called?
kinetic energy
potential energy
bond energy
internal energy
D
A 0.10 kg piece of copper at an initial temperature of 95°C is dropped into 0.20 kg of water contained in a 0.28 kg aluminum calorimeter. The water and calorimeter are initially at 15°C. What is the final temperature of the system when it reaches equilibrium?
mw = 0.20kg
cp,w = 4186J/kg°C
mAl = 0.28kg
cp,Al = 899J/kg°C
Tw = TAl = 15°C
mCu = 0.10kg
cp,Cu = 387J/kg°C
TCu = 95°C
18
Thank you!
You're welcome! I'm happy to help.
To improve the efficiency of a heat engine, one way is to increase Qh, which stands for the heat input at the high-temperature reservoir. By increasing the amount of heat transferred into the engine, more work can be generated, resulting in higher efficiency. So, the correct option to improve the efficiency of a heat engine is A - increase Qh.
The energies associated with atomic motion are called internal energy. Internal energy includes both the kinetic energy and potential energy of the atoms and molecules in a substance. So, the correct option for the energy associated with atomic motion is D - internal energy.
Now, let's move on to the equilibrium temperature question. We have a system involving a piece of copper, water, and an aluminum calorimeter. Initially, the copper is at 95°C, the water and calorimeter are at 15°C, and we need to find the final equilibrium temperature of the system.
To solve this problem, we need to consider the heat exchange that occurs until the system reaches equilibrium. The heat transfer from the copper to the water can be calculated using 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.
First, calculate the heat transfer from the copper to the water and calorimeter:
Qcopper = mcopper * cp,copper * (Tfinal - Tcopper)
Qwater = mwater * cp,water * (Tfinal - Twater)
Qcalorimeter = mcalorimeter * cp,calorimeter * (Tfinal - Tcalorimeter)
Since the system is in equilibrium, the total heat gained by the water and calorimeter must equal the heat lost by the copper:
Qwater + Qcalorimeter = Qcopper
Substitute the given values into the equations:
mwater * cp,water * (Tfinal - Twater) + mcalorimeter * cp,calorimeter * (Tfinal - Tcalorimeter) = mcopper * cp,copper * (Tfinal - Tcopper)
Now, plug in the given values:
(0.20 kg) * (4186 J/kg°C) * (Tfinal - 15°C) + (0.28 kg) * (899 J/kg°C) * (Tfinal - 15°C) = (0.10 kg) * (387 J/kg°C) * (95°C - Tfinal)
Simplify and solve for Tfinal:
(836.8 J/°C) * Tfinal - (12552 J) + (251.72 J/°C) * Tfinal - (3775.8 J) = (38.7 J/°C) * Tfinal - (3,870.0 J)
(836.8 J/°C + 251.72 J/°C - 38.7 J/°C) * Tfinal = (12552 J + 3775.8 J - 3,870.0 J)
(1050.8 J/°C) * Tfinal = 9657.8 J
Tfinal = 9657.8 J / 1050.8 J/°C
Tfinal ≈ 9.18°C
Therefore, the final temperature of the system when it reaches equilibrium is approximately 9.18°C.
I hope this helps! Let me know if you have any more questions.