A metallic sphere cools from $50^{\circ} \mathrm{C}$ to $40^{\circ}$ in 300 s . If atmospheric temperature around is $20^{\circ} \mathrm{C}$, then the sphere's temperature after the next 5 minutes will be close to:

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Heat & Thermodynamics

A metallic sphere cools from

5 0^{\circ} C

4 0^{\circ}

in 300 s . If atmospheric temperature around is

2 0^{\circ} C

, then the sphere's temperature after the next 5 minutes will be close to:

Answer: (C)

By Newton's law of cooling the rate of fall of temperature is proportional to the excess over the surroundings. Over the first

300 s

\frac{50 - 40}{300} = k (\frac{50 + 40}{2} - 20) = 25 k

, so

k = \frac{1}{750} s^{- 1}

. Over the next

300 s

the sphere starts at

4 0^{\circ}

C and reaches

T

\frac{40 - T}{300} = \frac{1}{750} (\frac{40 + T}{2} - 20) = \frac{1}{750} \cdot \frac{T}{2}

. This reduces to

\frac{40 - T}{2} = \frac{T}{5}

, i.e.

200 - 5 T = 2 T

, giving

T = \frac{200}{6} \approx 3 3^{\circ}

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Single correct option