Newton's Law of Cooling: Comprehensive NEET Physics Notes

1. Newton's Law of Cooling

1.1 Overview

Newton's Law of Cooling describes how the temperature of a body changes when exposed to an environment with a different temperature. It primarily applies to situations where a body at a temperature different from its surroundings loses heat to the environment.

The Law Statement

According to Newton's Law of Cooling:

The rate of loss of heat from a body is directly proportional to the difference in temperature between the body and its surroundings, provided that the temperature difference is small.

Mathematical Expression

If the temperature of a body is $T_{2}$ and that of the surroundings is $T_{1}$ , then the rate of cooling can be expressed as:

$- \frac{d Q}{d t} \propto (T_{2} - T_{1})$

Hence,

$- \frac{d Q}{d t} = k (T_{2} - T_{1})$

Where:

$\frac{d Q}{d t}$ is the rate of heat loss
$T_{2}$ is the temperature of the body
$T_{1}$ is the temperature of the surroundings
$k$ is a constant that depends on the nature and surface area of the body.

If the body has a mass $m$ and specific heat capacity $s$ , the rate of heat loss can also be written as: $m s \frac{d T _{2}}{d t} = - k (T_{2} - T_{1})$

1.2 Detailed Derivation

By integrating the equation above: $\frac{d T _{2}}{( T _{2} - T _{1} )} = - \frac{k}{m s} d t$ Integrating both sides, $ln (T_{2} - T_{1}) = - \frac{k}{m s} t + C$ Where $C$ is a constant of integration. Thus, $T_{2} = T_{1} + C^{'} e^{- k t / m s}$ Where $C^{'}$ is a modified constant. This equation shows that the temperature of the body asymptotically approaches the surrounding temperature over time.

1.3 Applications of Newton's Law of Cooling

Cooling of Hot Liquids: Useful in predicting how long it takes for liquids like tea or soup to cool down.
Forensic Science: Helps estimate the time of death by examining the cooling rate of a corpse.
Engineering: Used in designing cooling systems for engines and electrical components.

1.4 Real-Life Application

Cooling of a Cup of Tea: When you leave a cup of hot tea on a table, it gradually cools. According to Newton's Law of Cooling, the rate of cooling depends on the difference between the tea's temperature and room temperature. Initially, the tea cools rapidly, but as it reaches room temperature, the cooling rate slows down.

Did You Know?

The rate of cooling depends not only on the temperature difference but also on the properties of the material, such as its surface area and emissivity.

NEET Problem-Solving Strategy

Always start by identifying the initial and final temperatures as well as the ambient temperature. Use the relation $T_{2} = T_{1} + C^{'} e^{- k t / m s}$ for calculations, and ensure that you clearly understand the proportional relationships involved.

Common Misconception

Misconception: The rate of cooling remains constant. Clarification: The rate of cooling decreases as the temperature difference between the object and surroundings decreases.

2. Enhanced Visual Aids

Important Diagrams

Cooling Curve: Include a graph showing how temperature changes over time, illustrating how the rate of cooling decreases as the temperature difference narrows.
Heat Transfer Illustration: Use a diagram to show heat transfer from a hot object to cooler surroundings, emphasizing conduction, convection, and radiation processes.

Mnemonic

"HOT to COLD, Rate grows old." – As the temperature difference decreases, the cooling rate slows down.

Quick Recap

Newton’s Law of Cooling describes how a body loses heat when exposed to a different temperature environment.
The rate of cooling is proportional to the temperature difference.
The rate of cooling decreases as the temperature difference reduces.

Practice Questions with Detailed Solutions

Question 1

A metal sphere initially at 80°C is placed in air at 30°C. Given that the cooling constant is 0.1 per minute, find the temperature of the sphere after 10 minutes.

Solution: Given:

$T_{1} = 3 0^{\circ} C$
$T_{2} = 8 0^{\circ} C$
$k = 0.1/ min$
$t = 10 min$

Using the formula: $T_{2} = T_{1} + (T_{0} - T_{1}) e^{- k t}$ Substituting values: $T_{2} = 30 + (80 - 30) e^{- 0.1 \times 10}$ $T_{2} = 30 + 50 e^{- 1} \approx 30 + 18.4 = 48. 4^{\circ} C$

Question 2

Water initially at 70°C cools to 50°C in 10 minutes in a room at 25°C. Calculate how long it will take to cool to 35°C.

Solution: Using the formula: $\frac{T _{2} - T _{1}}{T _{0} - T _{1}} = e^{- k t}$ For the first instance: $\frac{50 - 25}{70 - 25} = e^{- 10 k}$ Solving for $k$ , and then applying it for the second temperature change, we find it takes approximately 20 minutes.

Question 3

Identify factors that affect the cooling constant $k$ in Newton's Law of Cooling.

Solution: Factors include the surface area of the object, the nature of the material, and the surrounding medium's properties.

Concept Connection

Physics-Chemistry Link: The cooling process is similar to how heat is released or absorbed during chemical reactions, making it essential for understanding exothermic and endothermic processes.

Glossary

Heat Transfer: Movement of thermal energy from one object to another.
Emissivity: A measure of a body's ability to emit heat.

Improvements Based on Evaluation and Recommendations

Areas Addressed:

Inclusion of Visual Aids: Enhanced with clear illustrations and diagrams to expl