Answer:
A)Qa=80 J
B)Qr= 100 J
Explanation:
Given that
W= 20 J
COP = 4
Heat rejected from cold reservoir = Qa
Heat exhausted to hot reservoir = Qr
The COP of refrigerator is given as
Qa= 4 x 20 J
Qa=80 J
By using first law of refrigerator
Qr= Qa + W
Qr= 80 + 20 J
Qr= 100 J
A)Qa=80 J
B)Qr= 100 J
Answer:
232 J/K
Explanation:
The amount of heat gained by the air = the amount of heat lost by the tea.
q_air = -q_tea
q = -mCΔT
q = -(0.250 kg) (4184 J/kg/ºC) (20.0ºC − 85.0ºC)
q = 68,000 J
The change in entropy is:
dS = dQ/T
Since the room temperature is constant (isothermal):
ΔS = ΔQ/T
Plug in values (remember to use absolute temperature):
ΔS = (68,000 J) / (293 K)
ΔS = 232 J/K
If a coin is dropped at a relatively low altitude, it's acceleration remains constant. However, if the coin is dropped at a very high altitude, air resistance will have a significant effect. The initial acceleration of the coin will be the greatest. As it falls down, air resistance will counteract the weight of the coin. So, the acceleration will decrease. Although the acceleration decreases, the coin still accelerates, that is why it falls faster. When the air resistance fully counters the weight of the coin, the acceleration will become zero and the coin will fall at a constant speed (terminal velocity). So, the answer should be, The acceleration decreases until it reaches 0. The closest answer is.
a. The acceleration decreases.
It forces the mercury to rise, being pushed up the tube by pressing down on the dish.
I hope I could help :)
