A metallic sphere cools from 50 ° C to 40 ° C in 300 s . If the room temperature is 20 ° C , then its temperature in the next 5 min will be

Question

A metallic sphere cools from 50 ° C to 40 ° C in 300 s . If the room temperature is 20 ° C , then its temperature in the next 5 min will be (A) 38° C (B) 33.3° C (C) 30° C (D) 36° C

Tardigrade Admin · Accepted Answer

According to Newton's law of cooling (θ 1 - θ 2/t)=K [(θ 1 - θ 2/t) - θ 0] (50 - 40/300)=K[(50 + 40/2) - 20] or K= (1/25 × 30) (40 - θ /300)=K[(40 + θ /2) - 20]=(K θ /2)=(θ /1500) Or 300 θ =60000-1500 θ , θ =(60000/1800)=33.3â

Q. A metallic sphere cools from $50^{\circ} C$ to $40^{\circ} C$ in $300 s$ . If the room temperature is $20^{\circ} C$ , then its temperature in the next $5 min$ will be

$3 8^{\circ} C$

$33. 3^{\circ} C$

$3 0^{\circ} C$

$3 6^{\circ} C$

Q. A metallic sphere cools from 50∘C to 40∘C in 300s . If the room temperature is 20∘C , then its temperature in the next 5min will be

38∘C

33.3∘C

30∘C

36∘C

According to Newton's law of cooling tθ1​−θ2​​=K[tθ1​−θ2​​−θ0​] 30050−40​=K[250+40​−20]orK=25×301​ 30040−θ​=K[240+θ​−20]=2Kθ​=1500θ​ Or 300θ=60000−1500θ , θ=180060000​=33.3℃

Q. A metallic sphere cools from $50^{\circ} C$ to $40^{\circ} C$ in $300 s$ . If the room temperature is $20^{\circ} C$ , then its temperature in the next $5 min$ will be

$3 8^{\circ} C$

$33. 3^{\circ} C$

$3 0^{\circ} C$

$3 6^{\circ} C$