A metallic sphere cools from 50ᵒC to 40ᵒ in 300 s. If atmospheric temperature around is 20ᵒC, then the sphere’s temperature after the next 5 minutes will be close to :

Solution

Physics sir · 4 steps

Watch the excess over 20°C: it shrinks by the same fraction in every equal time interval.

1. Track the excess. Newton's law says the cooling rate is proportional to T - T₀, the excess over the surroundings. So in equal time intervals the excess is multiplied by the same factor.

   (T₂ - T₀)/(T₁ - T₀) = e⁻ᵏᵗ

2. First 300 s. The excess falls from 50 - 20 = 30 °C to 40 - 20 = 20 °C. So in 300 s the excess is multiplied by 20/30 = 2/3.
3. Next 300 s. Five minutes is again 300 s, so the excess is multiplied by 2/3 once more: 20 × 2/3 = 13.3 °C. The averaged form of the law gives the same result, 40 - T = 0.2T.
4. So. T = 20 + 13.3 = 33.3 °C, which is closest to 33°C, option (c).

The idea to keep. In Newton's cooling, equal times mean equal fractions, not equal degrees. Find the ratio of the excess temperatures in one interval and apply it again for the next one.

Answer from the NTA final answer key. Solution by PrepWiser.