A cup of tea cools from 80° C to 60° C in one minute. The ambient temperature is 30° C. In cooling from 60° C to 50° C. It will take

Question

A cup of tea cools from 80° C to 60° C in one minute. The ambient temperature is 30° C. In cooling from 60° C to 50° C. It will take (A) 50 s (B) 90 s (C) 60 s (D) 48 s

Tardigrade Admin · Accepted Answer

Newton's law of cooling (Δ θ/Δ t)=k[θ-θ0] θ0= surrounding's temperature ⇒ (80-60/t)=k[(80+60/2)-30] ...(i) and (60-50/t)=k[(60+50/2)-30] ....(ii) ⇒ t=48 sec .

Q. A cup of tea cools from $8 0^{\circ} C$ to $6 0^{\circ} C$ in one minute. The ambient temperature is $3 0^{\circ} C$ . In cooling from $6 0^{\circ} C$ to $5 0^{\circ} C$ . It will take

50 s

90 s

60 s

48 s

Newton's law of cooling
$\frac{Δ θ}{Δ t} = k [θ - θ_{0}]$
$θ_{0} =$ surrounding's temperature
$\Rightarrow \frac{80 - 60}{t} = k [\frac{80 + 60}{2} - 30]$ ...(i)
and $\frac{60 - 50}{t} = k [\frac{60 + 50}{2} - 30]$ ....(ii)
$\Rightarrow t = 48 sec .$

Q. A cup of tea cools from 80∘C to 60∘C in one minute. The ambient temperature is 30∘C. In cooling from 60∘C to 50∘C. It will take