A cup of tea is placed at a table, and the room temperature is 35°C. The tea cools from 90°C to 70°C in 5 minutes. Which one among the following is the correct time taken for the tea to cool from 70°C to 50°C ?
In the first interval, the tea cools from 90°C to 70°C in t₁ = 5 minutes.
The temperature change is Δθ₁ = 90 – 70 = 20°C.
The average temperature in this interval is θ_avg₁ = (90 + 70) / 2 = 80°C.
The average temperature difference is ΔT₁ = θ_avg₁ – θ₀ = 80 – 35 = 45°C.
The average rate of cooling in this interval is approximately Rate₁ ≈ Δθ₁ / t₁ = 20°C / 5 min = 4°C/min.
By Newton’s Law of Cooling, Rate₁ ∝ ΔT₁, so 4 ∝ 45.
In the second interval, the tea cools from 70°C to 50°C. Let the time taken be t₂.
The temperature change is Δθ₂ = 70 – 50 = 20°C.
The average temperature in this interval is θ_avg₂ = (70 + 50) / 2 = 60°C.
The average temperature difference is ΔT₂ = θ_avg₂ – θ₀ = 60 – 35 = 25°C.
The average rate of cooling in this interval is approximately Rate₂ ≈ Δθ₂ / t₂ = 20°C / t₂.
By Newton’s Law of Cooling, Rate₂ ∝ ΔT₂, so (20/t₂) ∝ 25.
Taking the ratio of the rates:
Rate₁ / Rate₂ = ΔT₁ / ΔT₂
(4) / (20/t₂) = 45 / 25
4t₂ / 20 = 9 / 5
t₂ / 5 = 9 / 5
t₂ = 9 minutes.
The time taken for the tea to cool from 70°C to 50°C is 9 minutes.