A body cools from 75°C to 70°C in time t1, from 70°C to 65°C in time t2 and from 65°C to 60°C in time t3, then:

Question

A body cools from 75°C to 70°C in time t1, from 70°C to 65°C in time t2 and from 65°C to 60°C in time t3, then: (A)  t3 > t2 > t1  (B)  t1 > t2 > t3  (C)  t2 > t1= t3  (D)  t1 > t2 = t3

Tardigrade Admin · Accepted Answer

From Newtons law of cooling (dH/dt)=K( (θ 1+θ 2/2)-θ0 ) where θ 0 is temperature of surrounding, (θ1+θ2/2) the temperature of body. Hence, t3 > t2 > t1.

Q. A body cools from $7 5^{\circ} C$ to $7 0^{\circ} C$ in time $t_{1}$ , from $7 0^{\circ} C$ to $6 5^{\circ} C$ in time $t_{2}$ and from $6 5^{\circ} C$ to $6 0^{\circ} C$ in time $t_{3}$ , then:

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

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

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

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

From Newtons law of cooling
$\frac{d H}{d t} = K (\frac{θ _{1} + θ _{2}}{2} - θ_{0})$
where $θ_{0}$ is temperature of surrounding,
$\frac{θ _{1} + θ _{2}}{2}$
the temperature of body.
Hence, $t_{3} > t_{2} > t_{1} .$

Q. A body cools from 75∘C to 70∘C in time t1​, from 70∘C to 65∘C in time t2​ and from 65∘C to 60∘C in time t3​, then:

t3​>t2​>t1​

t1​>t2​>t3​

t2​>t1​=t3​

t1​>t2​=t3​