A cup of tea cools from 80° C to 60° C in one min. The ambient temperature is 30° C. In cooling from 60° C to 50° C, how much time (in s) will it take?

Question

A cup of tea cools from 80° C to 60° C in one min. The ambient temperature is 30° C. In cooling from 60° C to 50° C, how much time (in s) will it take? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

According to Newton's law of cooling (Δ θ/Δ t)=k[q-q0] q0= Surrounding's temperature ⇒ (80-60/60)=k[(80+60/2)-30] ...(i) and (60-50/t)=k[(60+50/2)-30] ...(ii) ⇒ t=48 s

Q. A cup of tea cools from $8 0^{\circ} C$ to $6 0^{\circ} C$ in one min. The ambient temperature is $3 0^{\circ} C$ . In cooling from $6 0^{\circ} C$ to $5 0^{\circ} C$ , how much time (in s) will it take?

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

Q. A cup of tea cools from 80∘C to 60∘C in one min. The ambient temperature is 30∘C. In cooling from 60∘C to 50∘C, how much time (in s) will it take?

According to Newton's law of cooling ΔtΔθ​=k[q−q0​] q0​= Surrounding's temperature ⇒6080−60​=k[280+60​−30] ...(i) and t60−50​=k[260+50​−30] ...(ii) ⇒t=48s

Q. A cup of tea cools from $8 0^{\circ} C$ to $6 0^{\circ} C$ in one min. The ambient temperature is $3 0^{\circ} C$ . In cooling from $6 0^{\circ} C$ to $5 0^{\circ} C$ , how much time (in s) will it take?

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