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Ice at –20°C is added tp 50 g of water at 40°C.When the temperature of the mixture reaches 0°C, it is found that 20 g of ice is still unmelted. The amount of ice added to the water was close to(Specific heat of water = 4.2 J/g/°C)Specific heat of Ice = 2.1 J/g/°CHeat of fusion of water at 0°C = 334 J/g)a)50 gb)40 gc)60 gd)100 gCorrect answer is option 'B'. 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 Ice at –20°C is added tp 50 g of water at 40°C.When the temperature of the mixture reaches 0°C, it is found that 20 g of ice is still unmelted. The amount of ice added to the water was close to(Specific heat of water = 4.2 J/g/°C)Specific heat of Ice = 2.1 J/g/°CHeat of fusion of water at 0°C = 334 J/g)a)50 gb)40 gc)60 gd)100 gCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Ice at –20°C is added tp 50 g of water at 40°C.When the temperature of the mixture reaches 0°C, it is found that 20 g of ice is still unmelted. The amount of ice added to the water was close to(Specific heat of water = 4.2 J/g/°C)Specific heat of Ice = 2.1 J/g/°CHeat of fusion of water at 0°C = 334 J/g)a)50 gb)40 gc)60 gd)100 gCorrect answer is option 'B'. Can you explain this answer?.

Solutions for Ice at –20°C is added tp 50 g of water at 40°C.When the temperature of the mixture reaches 0°C, it is found that 20 g of ice is still unmelted. The amount of ice added to the water was close to(Specific heat of water = 4.2 J/g/°C)Specific heat of Ice = 2.1 J/g/°CHeat of fusion of water at 0°C = 334 J/g)a)50 gb)40 gc)60 gd)100 gCorrect answer is option 'B'. 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 Ice at –20°C is added tp 50 g of water at 40°C.When the temperature of the mixture reaches 0°C, it is found that 20 g of ice is still unmelted. The amount of ice added to the water was close to(Specific heat of water = 4.2 J/g/°C)Specific heat of Ice = 2.1 J/g/°CHeat of fusion of water at 0°C = 334 J/g)a)50 gb)40 gc)60 gd)100 gCorrect answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of Ice at –20°C is added tp 50 g of water at 40°C.When the temperature of the mixture reaches 0°C, it is found that 20 g of ice is still unmelted. The amount of ice added to the water was close to(Specific heat of water = 4.2 J/g/°C)Specific heat of Ice = 2.1 J/g/°CHeat of fusion of water at 0°C = 334 J/g)a)50 gb)40 gc)60 gd)100 gCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for Ice at –20°C is added tp 50 g of water at 40°C.When the temperature of the mixture reaches 0°C, it is found that 20 g of ice is still unmelted. The amount of ice added to the water was close to(Specific heat of water = 4.2 J/g/°C)Specific heat of Ice = 2.1 J/g/°CHeat of fusion of water at 0°C = 334 J/g)a)50 gb)40 gc)60 gd)100 gCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of Ice at –20°C is added tp 50 g of water at 40°C.When the temperature of the mixture reaches 0°C, it is found that 20 g of ice is still unmelted. The amount of ice added to the water was close to(Specific heat of water = 4.2 J/g/°C)Specific heat of Ice = 2.1 J/g/°CHeat of fusion of water at 0°C = 334 J/g)a)50 gb)40 gc)60 gd)100 gCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Ice at –20°C is added tp 50 g of water at 40°C.When the temperature of the mixture reaches 0°C, it is found that 20 g of ice is still unmelted. The amount of ice added to the water was close to(Specific heat of water = 4.2 J/g/°C)Specific heat of Ice = 2.1 J/g/°CHeat of fusion of water at 0°C = 334 J/g)a)50 gb)40 gc)60 gd)100 gCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice JEE tests.