Newton's Law of Cooling

A body cools from $80{ }^{\mathrm{o}}\mathrm{C}$ to $60{ }^{\mathrm{o}}\mathrm{C}$ in $5$ minutes. The temperature of the surroundings is $20{ }^{\mathrm{o}}\mathrm{C}$. The time it takes to cool from $60{ }^{\mathrm{o}}\mathrm{C}$ to $40{ }^{\mathrm{o}}\mathrm{C}$ is

A body cools in $7$ minutes from ${60}^{\mathrm{o}}\mathrm{C}$ to ${40}^{\mathrm{o}}\mathrm{C}$. The temperature of the surrounding is ${10}^{\mathrm{o}}\mathrm{C}$. The temperature of the body after the next $7$ minutes will be

An object cools from $100°\mathrm{C}$ to $40°\mathrm{C}$ in $10$ minutes, when the surrounding temperature is $10°\mathrm{C}$. Then the time taken by the object to cool from $70°\mathrm{C}$ to $20°\mathrm{C}$ is
[Take $\ln 2=0.7,\ln 3=1.1,\ln 6=1.8$]

A hot container takes $1$ min to cool from $95 °\mathrm{C}$ to $75 °\mathrm{C}$. The time it takes to cool from $74 °\mathrm{C}$ to $54 °\mathrm{C}$ is
[consider the room temperature as $30 °\mathrm{C}$]

In a murder investigation a corpse was found by a detective at exactly at $8 \mathrm{pm}$. Being alert a detective is also measured the body temperature and found is to be $70°\mathrm{F}$. Two hours later, the detective measured the body temperature again and found to be $60°\mathrm{F}$. If the room temperature $50°\mathrm{F}$ and assuming that the body temperature of the person before death is $98.6°\mathrm{F}$, at what time did the murder occur? $\left[\log \left(2.43\right)=0.88789, \log \left(0.5\right)=-0.69315\right]$

The temperature of a cylindrical conductor starts rising when connected across a battery. The conductor radiates to the surrounding from it's curved surface following Newton's law of cooling and finally it attains a steady state temperature, its temperature exceeding the ambient temperature by $\Delta \mathrm{\theta }=9°\mathrm{C}.$ If the length of the conductor is reduced to $\left(\frac{3}{4}\right)$ times of the original length, keeping other conditions unchanged, the steady state temperature $\left({}^{\circ }\mathrm{C}\right)$ of the conductor exceeds the ambient temperature by

A body, placed in vacuum, starts cooling from the initial temperature of $T_0 \mathrm{K}$. Let $\Delta t$ be the time required to reduce the temperature to $T_0/2 \mathrm{K}$. Then, the approximate time needed to cool the body from $T_0 \mathrm{K}$ to $T_0/3 \mathrm{K}$ is

While finding the specific heat of a liquid, why the calorimeter is kept in a double walled chamber?

What are the limitations of the Newton's cooling law?

What is the principle of Newton's law of cooling?
 

What is the aim of verification of Newton's law of cooling?

What does a cooling curve show?

What does rate of cooling depend on?

How do you calculate cooling curve?

To do an experimental verification of Newton’s law of cooling, we need to take the ‘temperature’ vs ‘time’ readings, of a hot body, kept in given surroundings. An appropriate choice, of the ‘hot body’ and the ‘surroundings' would then be

Newton's law of cooling holds good provided the temperature difference between body and surrounding is

A body cools down from $52.5°\mathrm{C}$ to $47.5°\mathrm{C}$ in 5 minutes and from $47.5°\mathrm{C}$ to $42.5°\mathrm{C}$ in $7.5$ minute. Then the temperature of the surroundings is

A brass boiler has a base area $0.15{\mathrm{m}}^2$ and thickness $1.0\mathrm{cm}.$ It boils water at the rate of $6.0\mathrm{kg}/\min$ when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass:

A body cools from $80°\mathrm{C}$ to $50°\mathrm{C}$ In 5 minutes. Calculate the time it takes to cool from $60°\mathrm{C}$ to $30°\mathrm{C}$. The temperature of the surroundings is $20°\mathrm{C}$.

Certain quantity of water cools from $70 °\mathrm{C}$ to $60 °\mathrm{C}$ in the first 5 minutes and to $54 °\mathrm{C}$ in the next 5 minutes. The temperature of the surroundings is ;