Question

Explain the following :- 1) Hot tea cools rapidly when poured into the saucer from the cup b) temperature of a hot liquid falls rapidly in the beginning but slowly afterwards c) a hot liquid cools faster if outer surface of the container is blackened.

Answer 1

hey friend......

question:- Explain the following :- 1) Hot tea cools rapidly when poured into the saucer from the cup b) temperature of a hot liquid falls rapidly in the beginning but slowly afterwards c) a hot liquid cools faster if outer surface of the container is blackened.

Answer :- Answer is in this attachment

Answer 2

Hi

Here is your answer,

A) As surface area increases on pouring hot tea in saucer from the cup and the rate of loss of heat is directly proportional to surface area of the radiating surface, so the tea will cool faster in the saucer.

B) Temperature of a hot liquid falls exponentially in accordance with Netwon's law of cooling. In the other words, i can say that rate of cooling is directly proportional to the temperature difference between hot liquid and the surrounding. It is due to this reason that a hot liquid cools rapidly in the beginning but slowly afterwards.

C) When outer surfaces of container is blackened, the surface becomes good emitter of heat and so the hot liquid in it cools faster.


IN THE SECOND QUESTION I HAVE MENTIONED NEWTON'S LAW OF COOLING !

Let me tell you about Netwon's law of cooling 

Netwon's law of cooling states that the rate of cooling of a body is directly proportional to the temperature difference between the body and it's surrounding, provided the temperature is small.

Rate of loss of heat α Temperature difference  between the body and its surrounding.


                                → - dQ/dt α (T - T₀)

∴ Rate of loss of heat, -dQ/dt = k ( T - T₀)



Hope it helps you !