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. Essentially, 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

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

d₁ = displacement of tern from arctic to antarctic = 11000 miles

d₂ = displacement of tern from antarctic to arctic =   - 11000 miles

d = total displacement for the journey

total displacement for the journey is given as

d = d₁ + d₂

d = 11000 + (-11000)

d = 0 miles

t₁ = time of travel from arctic to antarctic = 3 months = 3 x 30 days = 90 days

t₂ = time of travel from antarctic to arctic = 3 months = 3 x 30 days = 90 days

t = total time of travel

total time of travel is given as

t = t₁ + t₂

t = 90 + 90

t = 180 days

average velocity is given as

average velocity = total displacement/ total time

v = d/t

v = 0/180

v = 0 miles/day