A cup of coffee cools from 90 ° C to 80 ° C in t minutes, when the room temperature is 20 ° C. The time taken by a similar cup of coffee to cool from 80 ° C to 60 ° C at a room temperature same at 20 ° C is: (a) 5/13 t (b) 13/10 t (c) 13/5 t (d) 10/13 t

Answer: (c). Newton's law: (T₁ - T₂)/(t) = k((T₁ + T₂)/(2) - T₀). 90 → 80: 10/t = k(85 - 20) = 65k 80 → 60: 20/t' = k(70 - 20) = 50k Dividing: t'/t = 20/10×65/50 = 13/5 So the answer is (c) 13/5t.

A cup of coffee cools from 90 ° C to 80 ° C in t minutes, when the… | NEET 2021

Q19NEET 2021Medium

Newton's Law of Cooling

A cup of coffee cools from 90 $^\circ$ C to 80 $^\circ$ C in t minutes, when the room temperature is 20 $^\circ$ C. The time taken by a similar cup of coffee to cool from 80 $^\circ$ C to 60 $^\circ$ C at a room temperature same at 20 $^\circ$ C is:

Show answer and solution

Answer: (c)

Newton's law: $\frac{T_1 - T_2}{t} = k\left(\frac{T_1 + T_2}{2} - T_0\right)$ .
$90 \to 80$ : $\frac{10}{t} = k(85 - 20) = 65k$
$80 \to 60$ : $\frac{20}{t'} = k(70 - 20) = 50k$
Dividing: $\frac{t'}{t} = \frac{20}{10}\times\frac{65}{50} = \frac{13}{5}$
So the answer is (c) $\frac{13}{5}t$ .

Report a mistake

Newton's Law of Cooling question from NEET 2021

More questions on Newton's Law of Cooling