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Q2. A block of 250 g of Copper at 120 degrees * C and 50 g of ice at - 25 degrees * C are submerged into 1 It of water at room temperature (25 degrees * C) . Calculate the final temperature at thermal equilibrium. Specific heat capacity of copper, ice, and water are 0.215 cal^ C , 0.5 cal/ C. and 1 cal/^ C respectively and latent heat of ice is 80cal / g? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Q2. A block of 250 g of Copper at 120 degrees * C and 50 g of ice at - 25 degrees * C are submerged into 1 It of water at room temperature (25 degrees * C) . Calculate the final temperature at thermal equilibrium. Specific heat capacity of copper, ice, and water are 0.215 cal^ C , 0.5 cal/ C. and 1 cal/^ C respectively and latent heat of ice is 80cal / g? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Q2. A block of 250 g of Copper at 120 degrees * C and 50 g of ice at - 25 degrees * C are submerged into 1 It of water at room temperature (25 degrees * C) . Calculate the final temperature at thermal equilibrium. Specific heat capacity of copper, ice, and water are 0.215 cal^ C , 0.5 cal/ C. and 1 cal/^ C respectively and latent heat of ice is 80cal / g?.

Solutions for Q2. A block of 250 g of Copper at 120 degrees * C and 50 g of ice at - 25 degrees * C are submerged into 1 It of water at room temperature (25 degrees * C) . Calculate the final temperature at thermal equilibrium. Specific heat capacity of copper, ice, and water are 0.215 cal^ C , 0.5 cal/ C. and 1 cal/^ C respectively and latent heat of ice is 80cal / g? 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 Q2. A block of 250 g of Copper at 120 degrees * C and 50 g of ice at - 25 degrees * C are submerged into 1 It of water at room temperature (25 degrees * C) . Calculate the final temperature at thermal equilibrium. Specific heat capacity of copper, ice, and water are 0.215 cal^ C , 0.5 cal/ C. and 1 cal/^ C respectively and latent heat of ice is 80cal / g? defined & explained in the simplest way possible. Besides giving the explanation of Q2. A block of 250 g of Copper at 120 degrees * C and 50 g of ice at - 25 degrees * C are submerged into 1 It of water at room temperature (25 degrees * C) . Calculate the final temperature at thermal equilibrium. Specific heat capacity of copper, ice, and water are 0.215 cal^ C , 0.5 cal/ C. and 1 cal/^ C respectively and latent heat of ice is 80cal / g?, a detailed solution for Q2. A block of 250 g of Copper at 120 degrees * C and 50 g of ice at - 25 degrees * C are submerged into 1 It of water at room temperature (25 degrees * C) . Calculate the final temperature at thermal equilibrium. Specific heat capacity of copper, ice, and water are 0.215 cal^ C , 0.5 cal/ C. and 1 cal/^ C respectively and latent heat of ice is 80cal / g? has been provided alongside types of Q2. A block of 250 g of Copper at 120 degrees * C and 50 g of ice at - 25 degrees * C are submerged into 1 It of water at room temperature (25 degrees * C) . Calculate the final temperature at thermal equilibrium. Specific heat capacity of copper, ice, and water are 0.215 cal^ C , 0.5 cal/ C. and 1 cal/^ C respectively and latent heat of ice is 80cal / g? theory, EduRev gives you an ample number of questions to practice Q2. A block of 250 g of Copper at 120 degrees * C and 50 g of ice at - 25 degrees * C are submerged into 1 It of water at room temperature (25 degrees * C) . Calculate the final temperature at thermal equilibrium. Specific heat capacity of copper, ice, and water are 0.215 cal^ C , 0.5 cal/ C. and 1 cal/^ C respectively and latent heat of ice is 80cal / g? tests, examples and also practice JEE tests.