Radiation and Newton's Law of Cooling | JEE Main previous year questions with Solutions

Q 1 :

The temperature of a body in air falls from 40°C to 24°C in 4 minute. The temperature of the air is 16°C. The temperature of the body in the next 4 minutes will be [2025]

14/3°C
28/3°C
56/3°C
42/3°C

1

2

3

4

(3)

$\frac{T_{2} - T_{1}}{t} = K [T_{avg} - T_{s}]$

$T_{1} = 24 ° C; T_{2} = 40 ° C; t = 4, T_{s} = 16 ° C$

$\frac{40 - 24}{4} = K [32 - 16]$

$K = \frac{4}{16} = \frac{1}{4}$

Now $\frac{24 - T}{4} = K [\frac{T + 24}{2} - 16]$

$24 - T = \frac{T - 8}{2}$

$\frac{3 T}{2} = 28 \Rightarrow T = \frac{56}{3} ° C^{}$

Q 2 :

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 the similar cup of coffee to cool from 80°C to 60°C at the same room temperature is: [2025]

$\frac{13}{5} t$
$\frac{10}{13} t$
$\frac{13}{10} t$
$\frac{5}{13} t$

1

2

3

4

(1)

By using average form of Newton's law of cooling

$\frac{90 - 80}{t} = c (\frac{90 + 80}{2} - 20)$ ... (i)

and $\frac{80 - 60}{t_{1}} = c (\frac{80 + 60}{2} - 20)$ ... (ii)

From (i) and (ii)

$t_{1} = \frac{13}{5} t$

Q 3 :

A wire of length 10 cm and diameter 0.5 mm is used in a bulb. The temperature of the wire is 1727°C and power radiated by the wire is 94.2 W. Its emissivity is $\frac{x}{8}$ where x = ________.

(Given $σ = 6.0 \times 10^{- 8} {Wm}^{- 2} K^{- 4}, π = 3.14$ and assume that the emissivity of wire material is same at all wavelength.) [2025]

0%

(5)

L = 10 cm, d = 0.5 mm, T = 1727°C = 2000 K

Power, P = 94.2 W

$P = ε σ A T^{4}$

$94.2 = ε \times (6 \times 10^{- 8}) (π d L) {(2000)}^{4}$

$94.2 = ε \times (6 \times 10^{- 8}) (3.14) (0.5) (10^{- 3}) (10 \times 10^{- 2}) {(2000)}^{4}$

$ε = \frac{5}{8} \Rightarrow x = 5$

Q 4 :

Two spherical bodies of same materials having radii 0.2 m and 0.8 m are placed in same atmosphere. The temperature of the smaller body is 800 K and temperature of bigger body is 400 K. If the energy radiated from the smaller body is E, the energy radiated from the bigger body is (assume, effect of the surrounding to be negligible) [2025]

256 E
E
64 E
16 E

1

2

3

4

(2)

$\frac{d θ}{d t} = σ e {AT}^{4} \Rightarrow P \propto {AT}^{4}$

$= \frac{P_{smaller}}{P_{larger}} = \frac{{(0.2)}^{2} \times 800^{4}}{{(0.8)}^{2} \times 400^{4}} = \frac{1}{16} \times 16 = 1$

$∴ P_{larger} = P_{smaller}$

$∴$ Energy will be same.

Q 5 :

A bowl filled with very hot soup cools from 98°C to 86°C in 2 minutes when the room temperature is 22°C. How long will it take to cool from 75°C to 69°C? [2023]

2 minutes
1 minute
1.4 minutes
0.5 minute

1

2

3

4

(3)

$\frac{θ_{2} - θ_{1}}{t} = k (\frac{θ_{1} + θ_{2}}{2} - θ_{0})$

$\frac{98 - 86}{2 min} = k (\frac{98 + 86}{2} - 22)$ ...(i)

$\frac{75 - 69}{t_{min}} = k [\frac{75 + 69}{2} - 22]$ ...(ii)

$Solving (i) and (ii), we get t = 1.4 min$

Q 6 :

A body cools in 7 minutes from 60°C to 40°C. The temperature of the surrounding is 10°C. The temperature of the body after the next 7 minutes will be [2023]

32°C
30°C
28°C
34°C

1

2

3

4

(3)

$Using average rate of Newton's law of cooling$

$\frac{T_{1} - T_{2}}{t} = K (\frac{T_{1} + T_{2}}{2} - T_{s})$

$Given \frac{60 - 40}{7} = K (50 - 10)$ ...(i)

$\frac{40 - T}{7} = K (\frac{40 + T}{2} - 10)$ ...(ii)

$From (i) and (ii), T = 28 ° C$

Q 7 :

A body cools from 80°C to 60°C in 5 minutes. The temperature of the surrounding is 20°C. The time it takes to cool from 60°C to 40°C is [2023]

500 s
$\frac{25}{3}$ s
450 s
420 s

1

2

3

4

(1)

$Rate of cooling \propto Temperature difference$

$\frac{80 - 60}{5} = k {70 - 20}$ ...(i)

$\frac{60 - 40}{t} = k {50 - 20}$ ...(ii)

$\frac{4 t}{20} = \frac{50}{30}$

$t = \frac{25}{3} min = 500 sec$

$\Rightarrow t = 500 sec$

Q 8 :

A body cools from 60°C to 40°C in 6 minutes. If the temperature of the surroundings is 10°C, then after the next 6 minutes, its temperature will be _____ °C. [2023]

0%

(28)

$By average form of Newton's law of cooling$

$\frac{20}{6} = k (50 - 10)$ ...(i)

$\frac{40 - T}{6} = k (\frac{40 + T}{2} - 10)$ ...(ii)

$From equations (i) and (ii)$

$\frac{20}{40 - T} = \frac{40}{10 + \frac{T}{2}}$

$10 + \frac{T}{2} = 80 - 2 T$

$\frac{5 T}{2} = 70 \Rightarrow T = 28 ° C$

Topic Question Set

Q 1 :

The temperature of a body in air falls from 40°C to 24°C in 4 minute. The temperature of the air is 16°C. The temperature of the body in the next 4 minutes will be [2025]

Q 2 :

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 the similar cup of coffee to cool from 80°C to 60°C at the same room temperature is: [2025]

0%

Q 5 :

A bowl filled with very hot soup cools from 98°C to 86°C in 2 minutes when the room temperature is 22°C. How long will it take to cool from 75°C to 69°C? [2023]

Q 6 :

A body cools in 7 minutes from 60°C to 40°C. The temperature of the surrounding is 10°C. The temperature of the body after the next 7 minutes will be [2023]

Q 7 :

A body cools from 80°C to 60°C in 5 minutes. The temperature of the surrounding is 20°C. The time it takes to cool from 60°C to 40°C is [2023]

Q 8 :

A body cools from 60°C to 40°C in 6 minutes. If the temperature of the surroundings is 10°C, then after the next 6 minutes, its temperature will be _____ °C. [2023]

0%

Want Us to Email you About Special Offers & Updates?