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kykrilka [37]
3 years ago
11

Why does the hood of a car heat up after the car has been running for a while? A) According to the fourth law of thermodynamics,

the temperature of the other parts of the car increases due to the coolant used for the engine. B) According to the first law of thermodynamics, the hood of the car heats up using heat from the surroundings in-order to achieve thermal equilibrium with the engine. C) According to the second law of thermodynamics, not all energy from the burnt fuel is used to do work on the piston. It also produces heat which warms other parts of the car. D) According to the third law of thermodynamics, the increase in the velocity of the car changes the entropy of the tires. To balance this change, the temperature of the other parts is increased.
Physics
2 answers:
Svetlanka [38]3 years ago
6 0

Answer:

C) According to the second law of thermodynamics, not all energy from the burnt fuel is used to do work on the piston. It also produces heat which warms other parts of the car.  

Explanation:

A) According to the fourth law of thermodynamics, the temperature of the other parts of the car increases due to the coolant used for the engine.  

B) According to the first law of thermodynamics, the hood of the car heats up using heat from the surroundings in-order to achieve thermal equilibrium with the engine.  

C) According to the second law of thermodynamics, not all energy from the burnt fuel is used to do work on the piston. It also produces heat which warms other parts of the car.  

D) According to the third law of thermodynamics, the increase in the velocity of the car changes the entropy of the tires. To balance this change, the temperature of the other parts is increased.

gtnhenbr [62]3 years ago
3 0

Answer:

c

Explanation:

kuz i smrt

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Answer:

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8 0
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Water enters a baseboard radiator at 180 °F and at a flow rate of 2.0 gpm. Assuming the radiator releases heat into the room at
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Answer:

Temperature of water leaving the radiator = 160°F

Explanation:

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Volumetric flowrate = 2 gallons/min = 0.000126 m³/s; density of water = 1000 kg/m³

ṁ = 1000 × 0.000126 = 0.126 kg/s

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H = ṁcΔT = 5861.42

ΔT = 5861.42/(0.126 × 4200) = 11.08 K = 11.08°C

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8 0
3 years ago
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2 years ago
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