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

Question

A body cools from 60° C to 40° C in 6 minutes. If, temperature of surroundings is 10° C. Then, after the next 6 minutes, its temperature will be ° C. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

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 equation (i) and (ii)

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

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

\frac{5 T}{2} = 70 \Rightarrow T = 2 8^{\circ} C

Q. A body cools from 60∘C to 40∘C in 6 minutes. If, temperature of surroundings is 10∘C. Then, after the next 6 minutes, its temperature will be ___∘C.

By average form of Newton's law of cooling 620​=k(50−10).....(i) 640−T​=K(240+T​−10)....(ii) From equation (i) and (ii) 40−T20​=10+T/240​ 10+2T​=80−2T 25T​=70⇒T=28∘C

Q. A body cools from $6 0^{\circ} C$ to $4 0^{\circ} C$ in $6$ minutes. If, temperature of surroundings is $1 0^{\circ} C$ . Then, after the next $6$ minutes, its temperature will be ___ $^{\circ} C$ .

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 equation (i) and (ii)
$\frac{20}{40 - T} = \frac{40}{10 + T /2}$
$10 + \frac{T}{2} = 80 - 2 T$
$\frac{5 T}{2} = 70 \Rightarrow T = 2 8^{\circ} C$