A body cools from $60^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$ in 6 minutes. If, temperature of surroundings is $10^{\circ} \mathrm{C}$. Then, after the next 6 minutes, its temperature will be ______ ${}^{\circ} \mathrm{C}$.

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Heat & Thermodynamics

A body cools from

6 0^{\circ} C

4 0^{\circ} C

in 6 minutes. If, temperature of surroundings is

1 0^{\circ} C

. Then, after the next 6 minutes, its temperature will be ______

^{\circ} C

Answer: 28

Solution

Newton's law of cooling in average form reads

\frac{Δ T}{Δ t} = k (T_{a v g} - T_{s})

. The first interval gives

\frac{60 - 40}{6} = k (50 - 10)

, so

k = \frac{1}{12}

per minute. For the next six minutes, cooling from

4 0^{\circ}

C to

T

\frac{40 - T}{6} = \frac{1}{12} (\frac{40 + T}{2} - 10)

, which simplifies to

5 T = 140

. After the next six minutes the body is at

2 8^{\circ}

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Integer answer

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