
The height of a waterfall is 50 m. If \[g = 9.8m{s^{ - 2}}\], the difference between the temperature at the top and the bottom of the waterfall is
(A) \[{1.17^{\rm O}}C\]
(B) \[{2.17^{\rm O}}C\]
(C) \[{0.117^{\rm O}}C\]
(D) \[{1.43^{\rm O}}C\]
Answer
180k+ views
Hint: The potential energy of the water is converted into heat energy this relation between potential energy and heat energy can be as \[mgh = ms\Delta t\]. We will put all the values and find the change in the temperature of the water.
Complete step by step answer:
It is given in the question that the height of the waterfall is 50 m. Then we have to find the difference between the temperature at the top and the bottom of the waterfall.
Here the potential energy of the water is converted into heat energy this relation between potential energy and heat energy can be as \[mgh = ms\Delta t\]. Here m is the mass of the water, g is the gravitational force which is equal to \[9.8m{s^{ - 2}}\], h is the height of the waterfall s is the specific heat and \[\Delta t\] is the temperature change.
We know that \[1J = 4.2C\], so we get \[1000J = 4200C\].
So, we get s = 4200 C.
The relation between the potential energy and heat is \[mgh = ms\Delta t\].
On cancelling the similar terms from both sides, we get-
\[gh = s\Delta t\]
\[\Delta t = \dfrac{{gh}}{s}\]
On putting the values of g, h, and ‘s’ in \[\Delta t\] we get-
\[\Delta t = \dfrac{{9.8 \times 50}}{{4200}}\]
\[\Delta t = \dfrac{{490}}{{4200}}\]
\[\Delta t = {0.117^{\rm O}}C\]
Thus, the difference between the temperature at the top and the bottom of the waterfall is \[\Delta t = {0.117^{\rm O}}C\].
Therefore, option c is correct.
Additional information:
The specific heat is the amount of heat required to raise the temperature of the water by one-degree Celsius. Specific heat capacity can be used to identify an unknown substance. Because specific heat capacity is the physical property of a substance.
Note:
One can make a mistake that they may take the value of s as 4.2 but we have to take the value of s as 4200 because we are finding the change in temperature of the water and the unit of mass i.e., water is taken in kg.
Complete step by step answer:
It is given in the question that the height of the waterfall is 50 m. Then we have to find the difference between the temperature at the top and the bottom of the waterfall.
Here the potential energy of the water is converted into heat energy this relation between potential energy and heat energy can be as \[mgh = ms\Delta t\]. Here m is the mass of the water, g is the gravitational force which is equal to \[9.8m{s^{ - 2}}\], h is the height of the waterfall s is the specific heat and \[\Delta t\] is the temperature change.
We know that \[1J = 4.2C\], so we get \[1000J = 4200C\].
So, we get s = 4200 C.
The relation between the potential energy and heat is \[mgh = ms\Delta t\].
On cancelling the similar terms from both sides, we get-
\[gh = s\Delta t\]
\[\Delta t = \dfrac{{gh}}{s}\]
On putting the values of g, h, and ‘s’ in \[\Delta t\] we get-
\[\Delta t = \dfrac{{9.8 \times 50}}{{4200}}\]
\[\Delta t = \dfrac{{490}}{{4200}}\]
\[\Delta t = {0.117^{\rm O}}C\]
Thus, the difference between the temperature at the top and the bottom of the waterfall is \[\Delta t = {0.117^{\rm O}}C\].
Therefore, option c is correct.
Additional information:
The specific heat is the amount of heat required to raise the temperature of the water by one-degree Celsius. Specific heat capacity can be used to identify an unknown substance. Because specific heat capacity is the physical property of a substance.
Note:
One can make a mistake that they may take the value of s as 4.2 but we have to take the value of s as 4200 because we are finding the change in temperature of the water and the unit of mass i.e., water is taken in kg.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs

Difference Between Mass and Weight

JEE Main 2023 April 13 Shift 2 Question Paper with Answer Key

JEE Main 2023 April 11 Shift 2 Question Paper with Answer Key

JEE Main 2023 April 10 Shift 2 Question Paper with Answer Key

JEE Main 2023 (April 8th Shift 2) Physics Question Paper with Answer Key

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

What is Hybridisation in Chemistry?

Other Pages
NCERT Solutions For Class 11 Physics Chapter 2 Motion In A Straight Line - 2025-26

NCERT Solutions For Class 11 Physics Chapter 1 Units and Measurements - 2025-26

NCERT Solutions For Class 11 Physics Chapter 3 Motion In A Plane - 2025-26

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

Motion in a Straight Line Class 11 Physics Chapter 2 CBSE Notes - 2025-26
