
In Joule’s Calorimeter experiment, when a current of $1$ Ampere is passed through a coil for a known interval of time ‘t’, the temperature of water increases from $30^\circ C$to $33^\circ C$. When a current $2A$ is passed through the same coil placed in the same quantity of water and for the same time, the temperature of water increases from $30^\circ C$ to,
$A.{\text{ }}33^\circ C$
$B.{\text{ }}36^\circ C$
$C.{\text{ }}39^\circ C$
$D.{\text{ }}42^\circ C$
Answer
522.3k+ views
Hint:A Calorimeter is a device used to measure the amount of heat involved in a chemical or physical process. For example, when an exothermic reaction occurs in a calorimeter, the heat produced by the reaction is absorbed by the solution, which increases its temperature. The principle of Calorimeter indicates the law of conservation of energy.
Formulae Used: We will be using the relation of$1$) Heat is related to the rise in temperature$H = mcdT$ and $2$) Joule’s equation of electrical heating $H = {i^2}Rt$
Complete step by step answer:
When a resistor absorbs electrical energy, it dissipates the energy in the form of heat H. If the resistor is placed in the calorimeter, the amount of heat produced can be measured when it is absorbed in the calorimeter. The heat Q produces a temperature rise $dT$. The resistor coil is immersed in the water in a calorimeter. The heat Q produces a temperature rise $dT$. The heat is related to $dT$
The heat generated in water that resulted in the temperature of water increases from $30^\circ C$to $33^\circ C$ is given by,
$H = mcdT$
Where m is the mass of water
$dT = $ \[(33 - 30) = 3\] is the temperature difference.
Current passed through the coil, ${i_1} = 1A$
Therefore H is equal to ${i^2}Rt$
Where, i, R, and t are the current, resistance of the coil, and temperature respectively.
$ \Rightarrow {i^2}Rt = mcdT$
Here R, t, C, and m remain the same, but current increased to${i_2} = 2A$
Hence, $i_2^2 = mcd{T'}$
Dividing the two equations,
$\dfrac{{dT'}}{{dT}} = \dfrac{{{2^2}}}{{{1^2}}}$ \[(^\circ C)\]
$dT' = 12$
Therefore, the final temperature$ = 30 + 12 = 42^\circ C$
Option \[D)\] ${42^0}C$
Note:A Calorimeter is a device used to measure the amount of heat involved in a chemical or physical process. For example, when an exothermic reaction occurs in a calorimeter, the heat produced by the reaction is absorbed by the solution, which increases its temperature. The principle of Calorimeter indicates the law of conservation of energy.
Formulae Used: We will be using the relation of$1$) Heat is related to the rise in temperature$H = mcdT$ and $2$) Joule’s equation of electrical heating $H = {i^2}Rt$
Complete step by step answer:
When a resistor absorbs electrical energy, it dissipates the energy in the form of heat H. If the resistor is placed in the calorimeter, the amount of heat produced can be measured when it is absorbed in the calorimeter. The heat Q produces a temperature rise $dT$. The resistor coil is immersed in the water in a calorimeter. The heat Q produces a temperature rise $dT$. The heat is related to $dT$
The heat generated in water that resulted in the temperature of water increases from $30^\circ C$to $33^\circ C$ is given by,
$H = mcdT$
Where m is the mass of water
$dT = $ \[(33 - 30) = 3\] is the temperature difference.
Current passed through the coil, ${i_1} = 1A$
Therefore H is equal to ${i^2}Rt$
Where, i, R, and t are the current, resistance of the coil, and temperature respectively.
$ \Rightarrow {i^2}Rt = mcdT$
Here R, t, C, and m remain the same, but current increased to${i_2} = 2A$
Hence, $i_2^2 = mcd{T'}$
Dividing the two equations,
$\dfrac{{dT'}}{{dT}} = \dfrac{{{2^2}}}{{{1^2}}}$ \[(^\circ C)\]
$dT' = 12$
Therefore, the final temperature$ = 30 + 12 = 42^\circ C$
Option \[D)\] ${42^0}C$
Note:A Calorimeter is a device used to measure the amount of heat involved in a chemical or physical process. For example, when an exothermic reaction occurs in a calorimeter, the heat produced by the reaction is absorbed by the solution, which increases its temperature. The principle of Calorimeter indicates the law of conservation of energy.
Recently Updated Pages
Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

What is the opposite of entropy class 11 chemistry CBSE

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE
