A body cools from $75{}^\circ C$ to ${{65}^{\circ }}C$ in 5 minutes. If the room temperature is $25{}^\circ C$ then the temperature of the body at the end of next 5 minutes is:
1. $57{}^\circ C$
2. $55{}^\circ C$
3. $54{}^\circ C$
4. $53{}^\circ C$
Answer
300.6k+ views
Hint: We will use the concept of Newton's law of cooling and its mathematical representation in order to solve this equation. Here we have to find the temperature of the body at the end of the next five minutes.
Formula used:
$t=\frac{1}{k}\ln \left( \frac{{{\theta }_{2}}-{{\theta }_{n}}}{{{\theta }_{1}}-{{\theta }_{n}}} \right)$
Complete answer:
Newton’s law of cooling states that the rate of loss of heat from a body is directly proportional to the temperature difference between the body and its surroundings. That is, hot tea cools faster than warm tea. Newton’s law of cooling actually explains the rate of cooling.
From Newton’s law of cooling, we have the equation as
Time, $t=\frac{1}{k}\ln \left( \frac{{{\theta }_{2}}-{{\theta }_{n}}}{{{\theta }_{1}}-{{\theta }_{n}}} \right)$
Given in the question that a body cools from $75{}^\circ C$ to $65{}^\circ C$ in 5 minutes and the room temperature is $25{}^\circ C$
$5=\frac{1}{k}\ln \left( \frac{65-25}{75-25} \right)-(1)$
Let $\theta$ be the final temperature after the next five minutes. Then we have the equation as:
$5=\frac{1}{k}\ln \left( \frac{\theta -25}{65-25} \right)-(2)$
On dividing equation (2) by equation (1), we get:
$1=\frac{\ln \left( \frac{\theta -25}{40} \right)}{\ln \left( \frac{40}{50} \right)}$
Taking antilog and then solving the equation we get the temperature of the body in the next five minutes as:
Temperature, $\theta =57{}^\circ C$
Therefore, the answer is option (1)
Note: For the second case you should remember that the initial temperature in this case is $65{}^\circ C$as it is the temperature the body reaches after the first five minutes. K in the equation is just a positive constant which is the same in both the cases since the body under observation is the same for both cases. K depends on the area and nature of the surface of the body. This problem can also be solved in different ways.
Formula used:
$t=\frac{1}{k}\ln \left( \frac{{{\theta }_{2}}-{{\theta }_{n}}}{{{\theta }_{1}}-{{\theta }_{n}}} \right)$
Complete answer:
Newton’s law of cooling states that the rate of loss of heat from a body is directly proportional to the temperature difference between the body and its surroundings. That is, hot tea cools faster than warm tea. Newton’s law of cooling actually explains the rate of cooling.
From Newton’s law of cooling, we have the equation as
Time, $t=\frac{1}{k}\ln \left( \frac{{{\theta }_{2}}-{{\theta }_{n}}}{{{\theta }_{1}}-{{\theta }_{n}}} \right)$
Given in the question that a body cools from $75{}^\circ C$ to $65{}^\circ C$ in 5 minutes and the room temperature is $25{}^\circ C$
$5=\frac{1}{k}\ln \left( \frac{65-25}{75-25} \right)-(1)$
Let $\theta$ be the final temperature after the next five minutes. Then we have the equation as:
$5=\frac{1}{k}\ln \left( \frac{\theta -25}{65-25} \right)-(2)$
On dividing equation (2) by equation (1), we get:
$1=\frac{\ln \left( \frac{\theta -25}{40} \right)}{\ln \left( \frac{40}{50} \right)}$
Taking antilog and then solving the equation we get the temperature of the body in the next five minutes as:
Temperature, $\theta =57{}^\circ C$
Therefore, the answer is option (1)
Note: For the second case you should remember that the initial temperature in this case is $65{}^\circ C$as it is the temperature the body reaches after the first five minutes. K in the equation is just a positive constant which is the same in both the cases since the body under observation is the same for both cases. K depends on the area and nature of the surface of the body. This problem can also be solved in different ways.
Recently Updated Pages
If the magnetizing field on a ferromagnetic material class 12 physics JEE_Main

A point charge is placed at the corner of a cube The class 12 physics JEE_Main

What changes occur if the monochromatic light used class 12 physics JEE_Main

The unit of specific conductance is A Ohm B Ohmmetre class 12 physics JEE_Main

Which lens is used in magnifying glass A Concave lens class 12 physics JEE_Main

Sir C V Raman won the Nobel Prize in which year A 1928 class 12 physics JEE_Main

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

Understanding Atomic Structure for Beginners

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
CBSE Class 12 Physics Question Paper 2026: Download SET-wise PDF with Answer Key & Analysis

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

Understanding Uniform Acceleration in Physics

Hybridisation in Chemistry – Concept, Types & Applications

Effective Nuclear Charge for JEE

Understanding Elastic Collisions in Two Dimensions

