CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

In a surrounding medium of temperature of 1$$0^{\circ}C$$, a body takes $$7$$ min for a fall of temperature from $$60^{\circ}C$$ to $$40^{\circ}C$$. In what time the temperature of the body will fall from $$40^{\circ}C$$ to $$28^{\circ}C$$?


A
7 min
loader
B
11 min
loader
C
14 min
loader
D
21 min
loader

Solution

The correct option is B 7 min
Newton’s Law of Cooling states that the rate fall of temperature of the body is proportional to the difference between the temperature of the body and that of the surrounding medium.
Thus $$\dfrac{\theta_2-\theta_1}{t}=K(\dfrac{\theta_1+\theta_2}{2}-\theta_0)$$
$$\implies \dfrac{60-40}{7}=K(\dfrac{60+40}{2}-10)$$
and $$\dfrac{40-28}{t}=K(\dfrac{40+28}{2}-10)$$
Solving the two equations gives $$t=7min$$

Physics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image