Rate of cooling of two identical beakers of hot water Consider a beaker of some volume of hot water at some initial temp say $75$ degrees celsius, which are left to cool. If the the temperature of water reaches $65$ degrees celsius in $30$ seconds. This is a cooling rate of $0.33$ $\frac{°C}{s}$. Now will the cooling rate decrease over the next $30$ seconds? If yes then what is the correct physical explanation.
 A: Newton’s law of cooling says the rate of heat loss of a body is directly proportional to the temperature difference between the body and its environment.
Therefore, assuming the temperature of the environment  of your beaker is constant, as the temperature of the beaker decreases the rate of cooling decreases.
Hope this helps.
A: Yes, the cooling rate will decrease. A way to see that is considering the Fourier's law of the thermal conduction, who states that if you have two materials in thermal contact, the cooling rate depends by the surface of contact between them and (beyond other things) by the difference between the temperatures of the two materials. In the case of the beakers of water, we have that they are in contact with the atmosphere which is approximately at $20$°C. The surface of contact is all the surface of the beakers (they are immerse in the atmosphere plus maybe the table on which they stay but now it doesn't matter for understand the concept). So the initial $\Delta T_1$ between the water and the atmosphere is $75-20=55$°C, after $30$' the temperature of water is $65$°C while the atmosphere's one is always at $20$°C (the atmosphere is so big with respect to the beakers, the change of its temperature is such small that it can be neglected). Now, the $\Delta T_2$ is now $65-20=45$°C and so the cooling rate will be lower than before because it is proportional to the $\Delta T$. The idea is that if you want to warm up something, you can do it faster putting it on a plate that is a $200$°C than on an other that is a $100$°C, and that's because of the Fourier's law of thermal conduction.
