JEE PYQ: Thermodynamics Question 17
Question 17 - 2020 (03 Sep Shift 2)
A metallic sphere cools from 50$°$C to 40$°$C in 300 s. If atmospheric temperature around is 20$°$C, then the sphere’s temperature after the next 5 minutes will be close to :
(1) 31$°$C
(2) 33$°$C
(3) 28$°$C
(4) 35$°$C
Show Answer
Answer: (2)
Solution
From Newton’s Law of cooling,
$\frac{T_1 - T_2}{t} = K\left(\frac{T_1 + T_2}{2} - T_0\right)$
Here, $T_1 = 50°$C, $T_2 = 40°$C and $T_0 = 20°$C, $t = 600$S = 5 minutes
$\frac{50 - 40}{5 \text{ Min}} = K\left(\frac{50 + 40}{2} - 20\right)$ …(i)
Let $T$ be the temperature of sphere after next 5 minutes. Then
$\frac{40 - T}{5} = K\left(\frac{40 + T}{2} - 20\right)$ …(ii)
Dividing eqn. (ii) by (i):
$\frac{40 - T}{10} = \frac{40 + T - 40}{50 + 40 - 40} = \frac{T}{50}$
$40 - T = \frac{T}{5} \Rightarrow 200 - 5T = T$
$\therefore T = \frac{200}{6} = 33.3°$C
BETA
