The temperature of a body falls from 40 ° C to 36 ° C in 4 minutes , when placed in a constant ambient temperature of 20 ° C . The time it takes for the temperature to fall from 36 ° C to 32 ° C in the same ambient temperature is

Question

The temperature of a body falls from 40 ° C to 36 ° C in 4 minutes , when placed in a constant ambient temperature of 20 ° C . The time it takes for the temperature to fall from 36 ° C to 32 ° C in the same ambient temperature is (A) 5.1 min (B) 6.8 min (C) 9 min (D) 7.2 min

Tardigrade Admin · Accepted Answer

Let θ be instantaneous temperature, θ For 40 ° C arrow 36 ° C (36 - 40/4)=-k(.38-20.) ...(.i.) θ a v g=38 ° C For 36 ° C arrow 32 ° C (32 - 36/t)=-k(.34-20.) ...(.ii.) θ a v g=34 ° C ((. i .)/(. ii .)) Gives t=4((18/14))⇒ t=5.1 min

Q. The temperature of a body falls from $40^{\circ} C$ to $36^{\circ} C$ in $4 min u t es$ , when placed in a constant ambient temperature of $20^{\circ} C$ . The time it takes for the temperature to fall from $36^{\circ} C$ to $32^{\circ} C$ in the same ambient temperature is

$5.1 min$

$6.8 min$

$9 min$

$7.2 min$

Q. The temperature of a body falls from 40∘C to 36∘C in 4minutes , when placed in a constant ambient temperature of 20∘C . The time it takes for the temperature to fall from 36∘C to 32∘C in the same ambient temperature is

5.1min

6.8min

9min

7.2min

Let θ be instantaneous temperature, θ For 40∘C→36∘C 436−40​=−k(38−20)...(i) θavg​=38∘C For 36∘C→32∘C t32−36​=−k(34−20)...(ii) θavg​=34∘C (ii)(i)​ Gives t=4(1418​)⇒t=5.1min

Q. The temperature of a body falls from $40^{\circ} C$ to $36^{\circ} C$ in $4 min u t es$ , when placed in a constant ambient temperature of $20^{\circ} C$ . The time it takes for the temperature to fall from $36^{\circ} C$ to $32^{\circ} C$ in the same ambient temperature is

$5.1 min$

$6.8 min$

$9 min$

$7.2 min$