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A piece of ice (heat capacity = 2100 J kg-1 °C-1 and latent heat = 3.36 ×105J kg-1) of mass m grams is at
-5°C at atmospheric pressure. It is given 420 J of heat so that the ice starts melting. Finally when the
ice-water mixture is in equilibrium, it is found that 1 gm of ice has melted. Assuming there is no other heat
exchange in the process, the value of m is
    Correct answer is '8'. Can you explain this answer?
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    A piece of ice (heat capacity = 2100 J kg-1 °C-1 and latent heat =...
    420 = (m × 2100 × 5 + 1 × 3.36 × 105 )×10−3
    where m is in gm.
    View all questions of this test
    Most Upvoted Answer
    A piece of ice (heat capacity = 2100 J kg-1 °C-1 and latent heat =...
    The heat capacity of a substance is the amount of heat required to raise the temperature of a given mass of the substance by 1 degree Celsius. In this case, the heat capacity of the ice is given as 2100 J kg-1.

    This means that for every kilogram of ice, 2100 joules of heat energy is required to raise its temperature by 1 degree Celsius.

    For example, if we have 2 kilograms of ice and we want to raise its temperature by 5 degrees Celsius, we can calculate the amount of heat required using the heat capacity:

    Heat energy = Heat capacity * Mass * Temperature change
    Heat energy = 2100 J kg-1 * 2 kg * 5 °C
    Heat energy = 21000 J

    Therefore, 21000 joules of heat energy would be required to raise the temperature of 2 kilograms of ice by 5 degrees Celsius.
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    A piece of ice (heat capacity = 2100 J kg-1 °C-1 and latent heat = 3.36 ×105J kg-1) of mass m grams is at-5°C at atmospheric pressure. It is given 420 J of heat so that the ice starts melting. Finally when theice-water mixture is in equilibrium, it is found that 1 gm of ice has melted. Assuming there is no other heatexchange in the process, the value of m isCorrect answer is '8'. Can you explain this answer?
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    A piece of ice (heat capacity = 2100 J kg-1 °C-1 and latent heat = 3.36 ×105J kg-1) of mass m grams is at-5°C at atmospheric pressure. It is given 420 J of heat so that the ice starts melting. Finally when theice-water mixture is in equilibrium, it is found that 1 gm of ice has melted. Assuming there is no other heatexchange in the process, the value of m isCorrect answer is '8'. 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 piece of ice (heat capacity = 2100 J kg-1 °C-1 and latent heat = 3.36 ×105J kg-1) of mass m grams is at-5°C at atmospheric pressure. It is given 420 J of heat so that the ice starts melting. Finally when theice-water mixture is in equilibrium, it is found that 1 gm of ice has melted. Assuming there is no other heatexchange in the process, the value of m isCorrect answer is '8'. 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 piece of ice (heat capacity = 2100 J kg-1 °C-1 and latent heat = 3.36 ×105J kg-1) of mass m grams is at-5°C at atmospheric pressure. It is given 420 J of heat so that the ice starts melting. Finally when theice-water mixture is in equilibrium, it is found that 1 gm of ice has melted. Assuming there is no other heatexchange in the process, the value of m isCorrect answer is '8'. Can you explain this answer?.
    Solutions for A piece of ice (heat capacity = 2100 J kg-1 °C-1 and latent heat = 3.36 ×105J kg-1) of mass m grams is at-5°C at atmospheric pressure. It is given 420 J of heat so that the ice starts melting. Finally when theice-water mixture is in equilibrium, it is found that 1 gm of ice has melted. Assuming there is no other heatexchange in the process, the value of m isCorrect answer is '8'. 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.
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