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Leona [35]
3 years ago
6

You prepare tea with 0.250 kg of water at 85.0 ºC and let it cool down to room temperature (20.0 ºC) before drinking it. Essenti

ally, the cooling process is isothermal to the air in your kitchen. Calculate the entropy change of the air as the tea cools, assuming that all the heat lost by the water goes into the air
Physics
1 answer:
vladimir2022 [97]3 years ago
5 0

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

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a driver travels 135 km, east in 1.5 h, stops for 45 minutes for lunch, and then resumes driving for the next 2.0 h through a di
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Answer:

v = 98.75 km/h

Explanation:

Given,

The distance driver travels towards the east, d₁ = 135 km

The time period of the travel, t₁ = 1.5 h

The halting time, tₓ = 46 minutes

The distance driver travels towards the east, d₂ = 215 km

The time period of the travel, t₁ = 2 h

The average speed of the vehicle before stopping

                                    v₁ = d₁/t₁

                                        = 135/1.5

                                       = 90 km/h

The average speed of vehicle after stopping

                                    v₂ = d₂/t₂

                                         = 215/2

                                        = 107.5 km/h

The total average velocity of the driver

                                      v = (v₁ +v₂) /2

                                         = (90 + 107.5)/2

                                         = 98.75 km/h

Hence, the average velocity of the driver, v = 98.75 km/h

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Answer:

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2 years ago
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