Question Description
A 0.60 kg sample of water and a sample of ice are placed in two compartments A and B that are separated by a conducting wall, in a thermally insulated container. The rate of heat transfer from the water to the ice through the conducting wall is constant P, until thermal equilibrium is reached. The temperature T of the liquid water and the ice are given in graph as functions of time t. Temperature of the each compartment remain homogeneous during whole heat transfer process. Given specific heat of ice = 2100 J/kg-K Given specific heat of water = 4200 J/kg-K Latent heat of fusion of ice = 3.3 × 105 J/kgQ.The mass of the ice formed due to conversion from the water till thermal equilibrium is reached, is equal toa)0.12 kgb)0.15 kgc)0.25 kgd)0.40 kgCorrect 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 A 0.60 kg sample of water and a sample of ice are placed in two compartments A and B that are separated by a conducting wall, in a thermally insulated container. The rate of heat transfer from the water to the ice through the conducting wall is constant P, until thermal equilibrium is reached. The temperature T of the liquid water and the ice are given in graph as functions of time t. Temperature of the each compartment remain homogeneous during whole heat transfer process. Given specific heat of ice = 2100 J/kg-K Given specific heat of water = 4200 J/kg-K Latent heat of fusion of ice = 3.3 × 105 J/kgQ.The mass of the ice formed due to conversion from the water till thermal equilibrium is reached, is equal toa)0.12 kgb)0.15 kgc)0.25 kgd)0.40 kgCorrect 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 A 0.60 kg sample of water and a sample of ice are placed in two compartments A and B that are separated by a conducting wall, in a thermally insulated container. The rate of heat transfer from the water to the ice through the conducting wall is constant P, until thermal equilibrium is reached. The temperature T of the liquid water and the ice are given in graph as functions of time t. Temperature of the each compartment remain homogeneous during whole heat transfer process. Given specific heat of ice = 2100 J/kg-K Given specific heat of water = 4200 J/kg-K Latent heat of fusion of ice = 3.3 × 105 J/kgQ.The mass of the ice formed due to conversion from the water till thermal equilibrium is reached, is equal toa)0.12 kgb)0.15 kgc)0.25 kgd)0.40 kgCorrect answer is option 'B'. Can you explain this answer?.

Solutions for A 0.60 kg sample of water and a sample of ice are placed in two compartments A and B that are separated by a conducting wall, in a thermally insulated container. The rate of heat transfer from the water to the ice through the conducting wall is constant P, until thermal equilibrium is reached. The temperature T of the liquid water and the ice are given in graph as functions of time t. Temperature of the each compartment remain homogeneous during whole heat transfer process. Given specific heat of ice = 2100 J/kg-K Given specific heat of water = 4200 J/kg-K Latent heat of fusion of ice = 3.3 × 105 J/kgQ.The mass of the ice formed due to conversion from the water till thermal equilibrium is reached, is equal toa)0.12 kgb)0.15 kgc)0.25 kgd)0.40 kgCorrect 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 A 0.60 kg sample of water and a sample of ice are placed in two compartments A and B that are separated by a conducting wall, in a thermally insulated container. The rate of heat transfer from the water to the ice through the conducting wall is constant P, until thermal equilibrium is reached. The temperature T of the liquid water and the ice are given in graph as functions of time t. Temperature of the each compartment remain homogeneous during whole heat transfer process. Given specific heat of ice = 2100 J/kg-K Given specific heat of water = 4200 J/kg-K Latent heat of fusion of ice = 3.3 × 105 J/kgQ.The mass of the ice formed due to conversion from the water till thermal equilibrium is reached, is equal toa)0.12 kgb)0.15 kgc)0.25 kgd)0.40 kgCorrect answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of A 0.60 kg sample of water and a sample of ice are placed in two compartments A and B that are separated by a conducting wall, in a thermally insulated container. The rate of heat transfer from the water to the ice through the conducting wall is constant P, until thermal equilibrium is reached. The temperature T of the liquid water and the ice are given in graph as functions of time t. Temperature of the each compartment remain homogeneous during whole heat transfer process. Given specific heat of ice = 2100 J/kg-K Given specific heat of water = 4200 J/kg-K Latent heat of fusion of ice = 3.3 × 105 J/kgQ.The mass of the ice formed due to conversion from the water till thermal equilibrium is reached, is equal toa)0.12 kgb)0.15 kgc)0.25 kgd)0.40 kgCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for A 0.60 kg sample of water and a sample of ice are placed in two compartments A and B that are separated by a conducting wall, in a thermally insulated container. The rate of heat transfer from the water to the ice through the conducting wall is constant P, until thermal equilibrium is reached. The temperature T of the liquid water and the ice are given in graph as functions of time t. Temperature of the each compartment remain homogeneous during whole heat transfer process. Given specific heat of ice = 2100 J/kg-K Given specific heat of water = 4200 J/kg-K Latent heat of fusion of ice = 3.3 × 105 J/kgQ.The mass of the ice formed due to conversion from the water till thermal equilibrium is reached, is equal toa)0.12 kgb)0.15 kgc)0.25 kgd)0.40 kgCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of A 0.60 kg sample of water and a sample of ice are placed in two compartments A and B that are separated by a conducting wall, in a thermally insulated container. The rate of heat transfer from the water to the ice through the conducting wall is constant P, until thermal equilibrium is reached. The temperature T of the liquid water and the ice are given in graph as functions of time t. Temperature of the each compartment remain homogeneous during whole heat transfer process. Given specific heat of ice = 2100 J/kg-K Given specific heat of water = 4200 J/kg-K Latent heat of fusion of ice = 3.3 × 105 J/kgQ.The mass of the ice formed due to conversion from the water till thermal equilibrium is reached, is equal toa)0.12 kgb)0.15 kgc)0.25 kgd)0.40 kgCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice A 0.60 kg sample of water and a sample of ice are placed in two compartments A and B that are separated by a conducting wall, in a thermally insulated container. The rate of heat transfer from the water to the ice through the conducting wall is constant P, until thermal equilibrium is reached. The temperature T of the liquid water and the ice are given in graph as functions of time t. Temperature of the each compartment remain homogeneous during whole heat transfer process. Given specific heat of ice = 2100 J/kg-K Given specific heat of water = 4200 J/kg-K Latent heat of fusion of ice = 3.3 × 105 J/kgQ.The mass of the ice formed due to conversion from the water till thermal equilibrium is reached, is equal toa)0.12 kgb)0.15 kgc)0.25 kgd)0.40 kgCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice JEE tests.