A tub of hot water cools from 80° C to 75° C in time t from 75° C to 70° C in time t 2, and from 70° C to 65° C in time t 3 then

Question

A tub of hot water cools from 80° C to 75° C in time t from 75° C to 70° C in time t 2, and from 70° C to 65° C in time t 3 then (A) t1>t2>t3 (B) t3>t2>t1 (C) t2>t1>t3 (D) t1>t2>t4

Tardigrade Admin · Accepted Answer

According to the Newton's law of cooling. Rate of cooling propto Mean temperature difference. (Δ T / t ) propto(( T 1+ T 2/2)- T 0 ) t propto (1/(( T 1+ T 2)2- T 0 )) So, t3>t2>t1

Q. A tub of hot water cools from $8 0^{\circ} C$ to $7 5^{\circ} C$ in time $t$ from $7 5^{\circ} C$ to $7 0^{\circ} C$ in time $t_{2}$ , and from $7 0^{\circ} C$ to $6 5^{\circ} C$ in time $t_{3}$ then

$t_{1} > t_{2} > t_{3}$

$t_{3} > t_{2} > t_{1}$

$t_{2} > t_{1} > t_{3}$

$t_{1} > t_{2} > t_{4}$

According to the Newton's law of cooling.
Rate of cooling $\propto$ Mean temperature difference.
$\frac{Δ T}{t} \propto (\frac{T _{1} + T _{2}}{2} - T_{0})$
$t \propto \frac{1}{( \frac{T _{1} + T _{2}}{2} - T _{0} )}$
So, $t_{3} > t_{2} > t_{1}$

Q. A tub of hot water cools from 80∘C to 75∘C in time t from 75∘C to 70∘C in time t2​, and from 70∘C to 65∘C in time t3​ then

t1​>t2​>t3​

t3​>t2​>t1​

t2​>t1​>t3​

t1​>t2​>t4​