1.00-L insulated bottle is full of tea at 90.08°C. You pour out one cup of tea and immediately screw the stopper back on the bottle. Make an order-of-magnitude estimate of the change in temperature of the tea remaining in the bottle that results from the admission of air at room temperature. State the quantities you take as data and the values you measure or estimate for them.

This is a calorimetry process where the heat given by the Te is absorbed by the air at room temperature (T₀ = 25ºC) with a specific heat of 1,009 J / kg ºC, we assume that the amount of Tea in the cup is V₀ = 100 ml. The bottle being thermally insulated does not intervene in the process

Qc = -Qb

M [tex]c_{e_Te}[/tex] (T₁ -[tex]T_{f}[/tex]) = m [tex]c_{e_air}[/tex] (T_{f}-T₀)

Where M is the mass of Tea that remains after taking out the cup, the density of Te is the density of water plus the solids dissolved in them, the approximate values are from 1020 to 1200 kg / m³, for this calculation we use 1100 kg / m³

ρ = m / V

V = 1000 -100 = 900 ml

V = 0.900 l (1 m3 / 1000 l) = 0.900 10⁻³ m³

V_air = 0.100 l = 0.1 10⁻³ m³

Tea Mass

M = ρ V_te

M = 1100 0.9 10⁻³

M = 0.990 kg

Air mass

m = ρ _air V_air

m = 1.225 0.1 10⁻³

m = 0.1225 10⁻³ kg

(m c_{e_air} + M c_{e_Te}) T_{f}. = M c_{e_Te} T1 - m c_{e_air} T₀

T_{f} = (M c_{e_Te} T₁ - m c_{e_air} T₀) / (m c_{e_air} + M c_{e_Te})

Let's calculate

T_{f} = (0.990 1100 90.08– 0.1225 10⁻³ 1.225 25) / (0.1225 10⁻³ 1.225 + 0.990 1100)

T_{f} = (98097.12 -3.75 10⁻³) / (0.15 10⁻³ +1089)

T_{f} = 98097.11 / 1089.0002

T_{f} = 90.07998 ° C

This temperature decrease is very small and cannot be measured