The temperature of hot water falls from 60^oC | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) According to Newton's Law of Cooling,the rate of cooling is given by: $\frac{d	heta}{dt} = K \left( \frac{	heta_1 + 	heta_2}{2} - 	heta_s ig

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(B) According to Newton's Law of Cooling,the rate of cooling is given by: $\frac{d	heta}{dt} = K \left( \frac{	heta_1 + 	heta_2}{2} - 	heta_s ight)$,where $	heta_s$ is the surrounding temperature.For the first interval ($60^oC$ to $50^oC$ in $10 \, min$):$\frac{60 - 50}{10} = K \left( \frac{60 + 50}{2} - 	heta_s ight)$$1 = K(55 - 	heta_s) \quad \dots(i)$For the second interval ($50^oC$ to $42^oC$ in $10 \, min$):$\frac{50 - 42}{10} = K \left( \frac{50 + 42}{2} - 	heta_s ight)$$0.8 = K(46 - 	heta_s) \quad \dots(ii)$Dividing equation $(i)$ by $(ii)$:$\frac{1}{0.8} = \frac{55 - 	heta_s}{46 - 	heta_s}$$1.25 = \frac{55 - 	heta_s}{46 - 	heta_s}$$1.25(46 - 	heta_s) = 55 - 	heta_s$$57.5 - 1.25	heta_s = 55 - 	heta_s$$2.5 = 0.25	heta_s$$	heta_s = 10^oC$.

The temperature of hot water falls from $60^oC$ to $50^oC$ in $10 \, min$. It takes another $10 \, min$ to fall from $50^oC$ to $42^oC$. Find the temperature of the surroundings in $^oC$.

Similar Questions

The temperature of a liquid in a beaker at time $t$ is $\theta(t)$ and the temperature of the surroundings is $\theta_0$. According to Newton's law of cooling,which of the following is the correct graph between $\log_e(\theta - \theta_0)$ and $t$?

There are two spherical balls $A$ and $B$ of the same material with the same surface,but the diameter of $A$ is half that of $B$. If $A$ and $B$ are heated to the same temperature and then allowed to cool,then

The instantaneous temperature difference between a cooling body and its surroundings,which obeys Newton's law of cooling,is $\theta$. Which of the following graphs represents the variation of $\ln \theta$ with time $t$?

Two metallic spheres $P$ and $Q$ are made of the same material and have the same smoothness,but the weight of $P$ is $8$ times that of $Q$. If the two are heated to the same temperature and left to cool,then the ratio of the rate of cooling of $Q$ to that of $P$ is:

Hot water cools from $80^{\circ} C$ to $60^{\circ} C$ in $1$ minute. How long will it take to cool from $60^{\circ} C$ to $50^{\circ} C$ (in $s$)? (Room temperature $= 30^{\circ} C$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module