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Asked in JEE Main 24th Jan 2nd Shift 2025 · Newton's law of cooling
Idea: Newton's law of cooling holds for a small excess over the surroundings; here every excess is modest, so work with the excess θ-θ₀ throughout.
In the first 4 minutes the excess falls from 40-16=24 °C to 24-16=8 °C, a factor of 3.
Equal intervals shrink the excess by equal factors, so in the next 4 minutes it falls from 8 °C to 8/3 °C.
The reading is then θ=16+8/3=(48+8)/3=(56)/3 °C, about 18.7 °C.
The average-temperature form gives the same figure: (24-θ)/4=k((24+θ)/2-16) with k=1/4 from the first interval leads again to 3θ=56.
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