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The specific heat of water = 4200 J kg-1K-1and the latent heat of ice = 3.4 × 105 J kg-1. 100 grams of ice at 0°C is placed in 200 g of water at 25°C. The amount of ice that will melt as the temperature of water reaches 0°C is close to (in grams) (Answer upto 1 decimal place)a)69.3b)63.8c)64.6d)61.7Correct answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The specific heat of water = 4200 J kg-1K-1and the latent heat of ice = 3.4 × 105 J kg-1. 100 grams of ice at 0°C is placed in 200 g of water at 25°C. The amount of ice that will melt as the temperature of water reaches 0°C is close to (in grams) (Answer upto 1 decimal place)a)69.3b)63.8c)64.6d)61.7Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The specific heat of water = 4200 J kg-1K-1and the latent heat of ice = 3.4 × 105 J kg-1. 100 grams of ice at 0°C is placed in 200 g of water at 25°C. The amount of ice that will melt as the temperature of water reaches 0°C is close to (in grams) (Answer upto 1 decimal place)a)69.3b)63.8c)64.6d)61.7Correct answer is option 'D'. Can you explain this answer?.

Solutions for The specific heat of water = 4200 J kg-1K-1and the latent heat of ice = 3.4 × 105 J kg-1. 100 grams of ice at 0°C is placed in 200 g of water at 25°C. The amount of ice that will melt as the temperature of water reaches 0°C is close to (in grams) (Answer upto 1 decimal place)a)69.3b)63.8c)64.6d)61.7Correct answer is option 'D'. Can you explain this answer? 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 specific heat of water = 4200 J kg-1K-1and the latent heat of ice = 3.4 × 105 J kg-1. 100 grams of ice at 0°C is placed in 200 g of water at 25°C. The amount of ice that will melt as the temperature of water reaches 0°C is close to (in grams) (Answer upto 1 decimal place)a)69.3b)63.8c)64.6d)61.7Correct answer is option 'D'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of The specific heat of water = 4200 J kg-1K-1and the latent heat of ice = 3.4 × 105 J kg-1. 100 grams of ice at 0°C is placed in 200 g of water at 25°C. The amount of ice that will melt as the temperature of water reaches 0°C is close to (in grams) (Answer upto 1 decimal place)a)69.3b)63.8c)64.6d)61.7Correct answer is option 'D'. Can you explain this answer?, a detailed solution for The specific heat of water = 4200 J kg-1K-1and the latent heat of ice = 3.4 × 105 J kg-1. 100 grams of ice at 0°C is placed in 200 g of water at 25°C. The amount of ice that will melt as the temperature of water reaches 0°C is close to (in grams) (Answer upto 1 decimal place)a)69.3b)63.8c)64.6d)61.7Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of The specific heat of water = 4200 J kg-1K-1and the latent heat of ice = 3.4 × 105 J kg-1. 100 grams of ice at 0°C is placed in 200 g of water at 25°C. The amount of ice that will melt as the temperature of water reaches 0°C is close to (in grams) (Answer upto 1 decimal place)a)69.3b)63.8c)64.6d)61.7Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice The specific heat of water = 4200 J kg-1K-1and the latent heat of ice = 3.4 × 105 J kg-1. 100 grams of ice at 0°C is placed in 200 g of water at 25°C. The amount of ice that will melt as the temperature of water reaches 0°C is close to (in grams) (Answer upto 1 decimal place)a)69.3b)63.8c)64.6d)61.7Correct answer is option 'D'. Can you explain this answer? tests, examples and also practice JEE tests.