Question 2
1 answer:
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Answer:
T_{f} = 90.07998 ° C
Explanation:
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 (T₁ -
) = m
(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