A body takes 4 min. to cool from 61° C to 59° C. If the temperature of the surroundings is 30° C, the time taken by the body to cool from 51° C to 49° C is :

Question

A body takes 4 min. to cool from 61° C to 59° C. If the temperature of the surroundings is 30° C, the time taken by the body to cool from 51° C to 49° C is : (A) 4 min (B) 3 min (C) 8 min (D) 6 min

Tardigrade Admin · Accepted Answer

(Δ T/Δ t)=K(Tt-Ts) Tt= average temp TS= surrounding temp. (61-59/4)=K((61+59/2)-30) ...(1) (51-49/t)=K((51+49/2)-30) ...(2) Divide (1) (2) (t/4)=(60-30/50-30)=(30/20) so, t=6 minutes

Q. A body takes $4 min$ . to cool from $6 1^{\circ} C$ to $5 9^{\circ} C$ . If the temperature of the surroundings is $3 0^{\circ} C$ , the time taken by the body to cool from $5 1^{\circ} C$ to $4 9^{\circ} C$ is :

4 min

3 \min

8 \min

6 \min

$\frac{Δ T}{Δ t} = K (T_{t} - T_{s})$
$T_{t} =$ average temp
$T_{S} =$ surrounding temp.
$\frac{61 - 59}{4} = K (\frac{61 + 59}{2} - 30)$ ...(1)
$\frac{51 - 49}{t} = K (\frac{51 + 49}{2} - 30)$ ...(2)
Divide (1) & (2)
$\frac{t}{4} = \frac{60 - 30}{50 - 30} = \frac{30}{20}$
so, $t = 6$ minutes

Q. A body takes 4min. to cool from 61∘C to 59∘C. If the temperature of the surroundings is 30∘C, the time taken by the body to cool from 51∘C to 49∘C is :