Hot water cools from 60° C to 50° C in the first 10 min and to 42° C in the next 10 min. The temperature of the surroundings is

Question

Hot water cools from 60° C to 50° C in the first 10 min and to 42° C in the next 10 min. The temperature of the surroundings is (A)  10° C  (B)  5° C  (C)  15° C  (D)  20° C

Tardigrade Admin · Accepted Answer

According to Newtons law of cooling (θ 1-θ 2/t)=K[ (θ 1+θ 2/2)-θ 0 ] In the first case ⇒ (60-50/10)=K[ (60+50/2)-θ 0 ] ⇒ 1=K(55-θ ) ...(i) In the second case ⇒ (50-42/10)=K[ (50+42/2)-θ ] ⇒ 0.8=K[46-θ ] ...(ii) Dividing Eq. (i) by Eq. (ii), we get (1/0.8)=(55-θ /46-θ ) or 46-θ =44-0.8θ θ =10° C

Q. Hot water cools from $60^{\circ} C$ to $50^{\circ} C$ in the first $10 min$ and to $42^{\circ} C$ in the next $10 min$ . The temperature of the surroundings is

$10^{\circ} C$

$5^{\circ} C$

$15^{\circ} C$

$20^{\circ} C$

$22^{\circ} C$

Q. Hot water cools from 60∘C to 50∘C in the first 10min and to 42∘C in the next 10min. The temperature of the surroundings is

10∘C

5∘C

15∘C

20∘C