Determine the law that is defined by the statement, “The rate of loss of heat $-\dfrac{dQ}{dt}$ of the body is directly proportional to the temperature difference $∆T=T_2-T_1$ of the body and surrounding”.
(A) Law of thermometry
(B) Newton’s law of cooling
(C) Law of calorimetry
(D) Zeroth law
Answer
624.3k+ views
Hint
According to the given statement, the rate of loss of heat of the body is directly proportional to the temperature difference of the body and surrounding which suggests that during the rate of loss of heat, the body will be cooling down. The cooling of the body will occur when there is a temperature difference between the body and the surrounding.
Complete step by step answer
Let us consider a body of mass $m$ at temperature $T$ whose specific heat is $s$. Also, the temperature of the surrounding is ${{\rm{T}}_0}$ such that ${\rm{T}} > {{\rm{T}}_0}$.
Since there will be loss of heat by the body, let the amount of heat lost by the body in time $dt$ be $dQ$.
Now, Newton’s law of cooling will be represented as,
$ - \dfrac{{{\rm{dQ}}}}{{{\rm{dt}}}} \propto \left( {{\rm{T}} - {{\rm{T}}_0}} \right)$
$ - \dfrac{{{\rm{dQ}}}}{{{\rm{dt}}}} = {\rm{k}}\left( {{\rm{T}} - {{\rm{T}}_0}} \right)$
Where $k$ is the proportionality constant. Also, the negative sign denotes the loss of heat during the transmission of heat.
Therefore, (B) Newton’s law of cooling is the required solution.
Note
The temperature always transfers from the body at a higher temperature to the body at a lower temperature. Newton’s law of cooling is applicable for a certain range of temperature differences. For Newton’s law of cooling, the temperature difference between the body and the surrounding should be small; not more than .
According to the given statement, the rate of loss of heat of the body is directly proportional to the temperature difference of the body and surrounding which suggests that during the rate of loss of heat, the body will be cooling down. The cooling of the body will occur when there is a temperature difference between the body and the surrounding.
Complete step by step answer
Let us consider a body of mass $m$ at temperature $T$ whose specific heat is $s$. Also, the temperature of the surrounding is ${{\rm{T}}_0}$ such that ${\rm{T}} > {{\rm{T}}_0}$.
Since there will be loss of heat by the body, let the amount of heat lost by the body in time $dt$ be $dQ$.
Now, Newton’s law of cooling will be represented as,
$ - \dfrac{{{\rm{dQ}}}}{{{\rm{dt}}}} \propto \left( {{\rm{T}} - {{\rm{T}}_0}} \right)$
$ - \dfrac{{{\rm{dQ}}}}{{{\rm{dt}}}} = {\rm{k}}\left( {{\rm{T}} - {{\rm{T}}_0}} \right)$
Where $k$ is the proportionality constant. Also, the negative sign denotes the loss of heat during the transmission of heat.
Therefore, (B) Newton’s law of cooling is the required solution.
Note
The temperature always transfers from the body at a higher temperature to the body at a lower temperature. Newton’s law of cooling is applicable for a certain range of temperature differences. For Newton’s law of cooling, the temperature difference between the body and the surrounding should be small; not more than .
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
What is the full form of CNG A Complete Natural Gas class 10 social science CBSE

In cricket, what is a "Yorker" designed to do?

What were the majoritarian measures taken in Sri Lanka class 10 social science CBSE

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE

10 examples of evaporation in daily life with explanations

Write a letter to the newspaper editor highlighting class 10 english CBSE

