JEE Exam > JEE Questions > The volume of mixture of ice and water 8s fou...
Start Learning for Free
The volume of mixture of ice and water 8s found to decrease by 0.132×10^-8 without change in temperature when 10 grams of metal at 100 celcius immersed in it the density of ice is 900kg/m^3 find specific heat in nearest integer if latent heat of ice is 336KJ/kg?
Most Upvoted Answer
The volume of mixture of ice and water 8s found to decrease by 0.132×1...
Given:
- Volume of mixture of ice and water decreases by 0.132×10^-8 m^3
- Mass of metal = 10 grams
- Temperature of metal = 100 degrees Celsius
- Density of ice = 900 kg/m^3
- Latent heat of ice = 336 kJ/kg
To Find:
Specific heat of the metal
Solution:
Step 1: Calculating the change in volume of the mixture
The change in volume of the mixture can be calculated using the formula:
Change in volume = Change in mass / Density
Given that the change in volume is -0.132×10^-8 m^3 and the density of ice is 900 kg/m^3, we can calculate the change in mass of the mixture.
Change in mass = Change in volume * Density
Change in mass = -0.132×10^-8 * 900 kg
Change in mass = -0.1188×10^-6 kg
Step 2: Calculating the heat absorbed by the mixture
The heat absorbed by the mixture can be calculated using the formula:
Heat absorbed = Change in mass * Latent heat of ice
Given that the change in mass is -0.1188×10^-6 kg and the latent heat of ice is 336 kJ/kg, we can calculate the heat absorbed by the mixture.
Heat absorbed = -0.1188×10^-6 * 336 kJ
Heat absorbed = -0.0399 kJ
Step 3: Calculating the heat gained by the metal
The heat gained by the metal can be calculated using the formula:
Heat gained = Mass of metal * Specific heat of metal * Change in temperature
Given that the mass of the metal is 10 grams, the temperature of the metal is 100 degrees Celsius, and there is no change in temperature, we can calculate the heat gained by the metal.
Heat gained = 10 grams * Specific heat of metal * 0 degrees Celsius
Heat gained = 0 kJ
Step 4: Equating the heat absorbed and heat gained
Since there is no change in temperature, the heat absorbed by the mixture must be equal to the heat gained by the metal.
-0.0399 kJ = 0 kJ
Step 5: Calculating the specific heat of the metal
From the equation above, we can conclude that the heat absorbed by the mixture is equal to the heat gained by the metal. Therefore,
-0.0399 kJ = 0 kJ
The specific heat of the metal is 0 kJ/kg°C.
Therefore, the specific heat of the metal is zero.
- Volume of mixture of ice and water decreases by 0.132×10^-8 m^3
- Mass of metal = 10 grams
- Temperature of metal = 100 degrees Celsius
- Density of ice = 900 kg/m^3
- Latent heat of ice = 336 kJ/kg
To Find:
Specific heat of the metal
Solution:
Step 1: Calculating the change in volume of the mixture
The change in volume of the mixture can be calculated using the formula:
Change in volume = Change in mass / Density
Given that the change in volume is -0.132×10^-8 m^3 and the density of ice is 900 kg/m^3, we can calculate the change in mass of the mixture.
Change in mass = Change in volume * Density
Change in mass = -0.132×10^-8 * 900 kg
Change in mass = -0.1188×10^-6 kg
Step 2: Calculating the heat absorbed by the mixture
The heat absorbed by the mixture can be calculated using the formula:
Heat absorbed = Change in mass * Latent heat of ice
Given that the change in mass is -0.1188×10^-6 kg and the latent heat of ice is 336 kJ/kg, we can calculate the heat absorbed by the mixture.
Heat absorbed = -0.1188×10^-6 * 336 kJ
Heat absorbed = -0.0399 kJ
Step 3: Calculating the heat gained by the metal
The heat gained by the metal can be calculated using the formula:
Heat gained = Mass of metal * Specific heat of metal * Change in temperature
Given that the mass of the metal is 10 grams, the temperature of the metal is 100 degrees Celsius, and there is no change in temperature, we can calculate the heat gained by the metal.
Heat gained = 10 grams * Specific heat of metal * 0 degrees Celsius
Heat gained = 0 kJ
Step 4: Equating the heat absorbed and heat gained
Since there is no change in temperature, the heat absorbed by the mixture must be equal to the heat gained by the metal.
-0.0399 kJ = 0 kJ
Step 5: Calculating the specific heat of the metal
From the equation above, we can conclude that the heat absorbed by the mixture is equal to the heat gained by the metal. Therefore,
-0.0399 kJ = 0 kJ
The specific heat of the metal is 0 kJ/kg°C.
Therefore, the specific heat of the metal is zero.
|
Explore Courses for JEE exam
|
|
Top Courses for JEEView all
Question Description
The volume of mixture of ice and water 8s found to decrease by 0.132×10^-8 without change in temperature when 10 grams of metal at 100 celcius immersed in it the density of ice is 900kg/m^3 find specific heat in nearest integer if latent heat of ice is 336KJ/kg? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The volume of mixture of ice and water 8s found to decrease by 0.132×10^-8 without change in temperature when 10 grams of metal at 100 celcius immersed in it the density of ice is 900kg/m^3 find specific heat in nearest integer if latent heat of ice is 336KJ/kg? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The volume of mixture of ice and water 8s found to decrease by 0.132×10^-8 without change in temperature when 10 grams of metal at 100 celcius immersed in it the density of ice is 900kg/m^3 find specific heat in nearest integer if latent heat of ice is 336KJ/kg?.
The volume of mixture of ice and water 8s found to decrease by 0.132×10^-8 without change in temperature when 10 grams of metal at 100 celcius immersed in it the density of ice is 900kg/m^3 find specific heat in nearest integer if latent heat of ice is 336KJ/kg? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The volume of mixture of ice and water 8s found to decrease by 0.132×10^-8 without change in temperature when 10 grams of metal at 100 celcius immersed in it the density of ice is 900kg/m^3 find specific heat in nearest integer if latent heat of ice is 336KJ/kg? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The volume of mixture of ice and water 8s found to decrease by 0.132×10^-8 without change in temperature when 10 grams of metal at 100 celcius immersed in it the density of ice is 900kg/m^3 find specific heat in nearest integer if latent heat of ice is 336KJ/kg?.
Solutions for The volume of mixture of ice and water 8s found to decrease by 0.132×10^-8 without change in temperature when 10 grams of metal at 100 celcius immersed in it the density of ice is 900kg/m^3 find specific heat in nearest integer if latent heat of ice is 336KJ/kg? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of The volume of mixture of ice and water 8s found to decrease by 0.132×10^-8 without change in temperature when 10 grams of metal at 100 celcius immersed in it the density of ice is 900kg/m^3 find specific heat in nearest integer if latent heat of ice is 336KJ/kg? defined & explained in the simplest way possible. Besides giving the explanation of
The volume of mixture of ice and water 8s found to decrease by 0.132×10^-8 without change in temperature when 10 grams of metal at 100 celcius immersed in it the density of ice is 900kg/m^3 find specific heat in nearest integer if latent heat of ice is 336KJ/kg?, a detailed solution for The volume of mixture of ice and water 8s found to decrease by 0.132×10^-8 without change in temperature when 10 grams of metal at 100 celcius immersed in it the density of ice is 900kg/m^3 find specific heat in nearest integer if latent heat of ice is 336KJ/kg? has been provided alongside types of The volume of mixture of ice and water 8s found to decrease by 0.132×10^-8 without change in temperature when 10 grams of metal at 100 celcius immersed in it the density of ice is 900kg/m^3 find specific heat in nearest integer if latent heat of ice is 336KJ/kg? theory, EduRev gives you an
ample number of questions to practice The volume of mixture of ice and water 8s found to decrease by 0.132×10^-8 without change in temperature when 10 grams of metal at 100 celcius immersed in it the density of ice is 900kg/m^3 find specific heat in nearest integer if latent heat of ice is 336KJ/kg? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Signup to solve all Doubts
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup