Question

A metal sphere cools at the rate of0.05 °C/s when its temperature is 70°C and at the rate of 0.025 °C/s whenits temperature is 50 °C. Determine thetemperature of the surroundings and findthe rate of cooling when the temperatureof the metal sphere is 40 °C.​

Answer 1

Therefore the surrounding temperature is 75°C

The rate of cooling when temperature of the metal is 40°C is 0.35°C/s

Explanation:

$(\frac{d\theta}{dt} )_1=0.05^\circ C$    when θ₁=70°C

$(\frac{d\theta}{dt})_2= 0.025^\circ C$    when θ₂= 50 °C

Newton's cooling law,

$(\frac{d\theta}{dt})=k(\theta-\theta_0)$

$\therefore \frac{(\frac{d\theta}{dt} )_1}{(\frac{d\theta}{dt})_2}= \frac{0.05}{0.025}$ $=\frac{(\theta_1-\theta_\circ)}{(\theta_2-\theta_\circ)}$ $=\frac{70-\theta_\circ}{50-\theta_\circ}$

$\Rightarrow \frac{70-\theta_\circ}{50-\theta_\circ} = \frac{0.05}{0.025}$

$\Rightarrow 70-\theta_\circ= 10-0.2\theta_\circ$

$\Rightarrow \theta_\circ= 75$

Then the value of K is

$k=\frac{0.05}{75-70}$ =0.01

Cooling rate at 40°C is

$\frac{d\theta}{dt} =0.01(75-40)$ =0.35°C/s

Therefore the surrounding temperature is 75°C

The rate of cooling when temperature of the metal is 40°C is 0.35°C/s

Answer 2

Answer:

0.0125

Explanation:

by newton's law of cooling,  

    0.05 = k ( 70 - s )

    0.025 = k ( 50 - s )

divide above equations...

    s = 30 -------first answer

 subs..s value in any one equation and get value of constant ( k )....

         the k = 0.05 / 40

         now temperature of metal sphere is 40

        dT/dt  =  0.05/40   ( 40 - s )

        dT/dt  =   0.05/40  ( 40 - 30)

 thus dT/dt  =   0.0125----second answer