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M grams of steam at 100°C is mixed with 200 g of ice at its melting point in a thermally insulated container. If it produces liquid water at 40°C [heat of vaporization of water is 540 cal/g and heat of fusion of ice is 80 cal/g], the value of M is _________.
Correct answer is '40'. Can you explain this answer?
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Most Upvoted Answer
M grams of steam at 100°C is mixed with 200 g of ice at its meltin...
°C is added to M grams of ice at 0°C. The final temperature of the mixture is 10°C. What is the value of M?
To solve this problem, we need to use the principle of conservation of energy. The heat gained by the ice equals the heat lost by the steam, and both are equal to the heat required to raise the temperature of the mixture from 0°C to 10°C.
The heat gained by the ice can be calculated using the formula:
Q_ice = m_ice * c_ice * (T_final - T_initial)
where m_ice is the mass of the ice, c_ice is the specific heat capacity of ice, and T_final and T_initial are the final and initial temperatures of the ice, respectively.
Similarly, the heat lost by the steam can be calculated using the formula:
Q_steam = m_steam * c_steam * (T_final - T_initial)
where m_steam is the mass of the steam, c_steam is the specific heat capacity of steam, and T_final and T_initial are the final and initial temperatures of the steam, respectively.
Since the final temperature of the mixture is 10°C, we have:
Q_ice = m_ice * c_ice * (10 - 0) = 10 * m_ice * c_ice
Q_steam = m_steam * c_steam * (10 - 100) = -90 * m_steam * c_steam
Since the heat gained by the ice is equal to the heat lost by the steam, we have:
10 * m_ice * c_ice = -90 * m_steam * c_steam
Dividing both sides of the equation by 10 * c_ice * c_steam, we get:
m_ice = -9 * m_steam
Since the masses of the steam and ice are equal (M grams), we can write:
M = -9 * M
Dividing both sides of the equation by M, we get:
1 = -9
This is not a valid solution, so there is no value of M that satisfies the given conditions.
To solve this problem, we need to use the principle of conservation of energy. The heat gained by the ice equals the heat lost by the steam, and both are equal to the heat required to raise the temperature of the mixture from 0°C to 10°C.
The heat gained by the ice can be calculated using the formula:
Q_ice = m_ice * c_ice * (T_final - T_initial)
where m_ice is the mass of the ice, c_ice is the specific heat capacity of ice, and T_final and T_initial are the final and initial temperatures of the ice, respectively.
Similarly, the heat lost by the steam can be calculated using the formula:
Q_steam = m_steam * c_steam * (T_final - T_initial)
where m_steam is the mass of the steam, c_steam is the specific heat capacity of steam, and T_final and T_initial are the final and initial temperatures of the steam, respectively.
Since the final temperature of the mixture is 10°C, we have:
Q_ice = m_ice * c_ice * (10 - 0) = 10 * m_ice * c_ice
Q_steam = m_steam * c_steam * (10 - 100) = -90 * m_steam * c_steam
Since the heat gained by the ice is equal to the heat lost by the steam, we have:
10 * m_ice * c_ice = -90 * m_steam * c_steam
Dividing both sides of the equation by 10 * c_ice * c_steam, we get:
m_ice = -9 * m_steam
Since the masses of the steam and ice are equal (M grams), we can write:
M = -9 * M
Dividing both sides of the equation by M, we get:
1 = -9
This is not a valid solution, so there is no value of M that satisfies the given conditions.
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Community Answer
M grams of steam at 100°C is mixed with 200 g of ice at its meltin...
M 540 + M × 1 × (100 - 40)
= 200 × 80 + 200 × 1 × 40
⇒ 600 M = 24000
⇒ M = 40
= 200 × 80 + 200 × 1 × 40
⇒ 600 M = 24000
⇒ M = 40
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M grams of steam at 100°C is mixed with 200 g of ice at its melting point in a thermally insulated container. If it produces liquid water at 40°C [heat of vaporization of water is 540 cal/g and heat of fusion of ice is 80 cal/g], the value of M is _________.Correct answer is '40'. Can you explain this answer?
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M grams of steam at 100°C is mixed with 200 g of ice at its melting point in a thermally insulated container. If it produces liquid water at 40°C [heat of vaporization of water is 540 cal/g and heat of fusion of ice is 80 cal/g], the value of M is _________.Correct answer is '40'. 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 M grams of steam at 100°C is mixed with 200 g of ice at its melting point in a thermally insulated container. If it produces liquid water at 40°C [heat of vaporization of water is 540 cal/g and heat of fusion of ice is 80 cal/g], the value of M is _________.Correct answer is '40'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for M grams of steam at 100°C is mixed with 200 g of ice at its melting point in a thermally insulated container. If it produces liquid water at 40°C [heat of vaporization of water is 540 cal/g and heat of fusion of ice is 80 cal/g], the value of M is _________.Correct answer is '40'. Can you explain this answer?.
M grams of steam at 100°C is mixed with 200 g of ice at its melting point in a thermally insulated container. If it produces liquid water at 40°C [heat of vaporization of water is 540 cal/g and heat of fusion of ice is 80 cal/g], the value of M is _________.Correct answer is '40'. 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 M grams of steam at 100°C is mixed with 200 g of ice at its melting point in a thermally insulated container. If it produces liquid water at 40°C [heat of vaporization of water is 540 cal/g and heat of fusion of ice is 80 cal/g], the value of M is _________.Correct answer is '40'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for M grams of steam at 100°C is mixed with 200 g of ice at its melting point in a thermally insulated container. If it produces liquid water at 40°C [heat of vaporization of water is 540 cal/g and heat of fusion of ice is 80 cal/g], the value of M is _________.Correct answer is '40'. Can you explain this answer?.
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M grams of steam at 100°C is mixed with 200 g of ice at its melting point in a thermally insulated container. If it produces liquid water at 40°C [heat of vaporization of water is 540 cal/g and heat of fusion of ice is 80 cal/g], the value of M is _________.Correct answer is '40'. Can you explain this answer?, a detailed solution for M grams of steam at 100°C is mixed with 200 g of ice at its melting point in a thermally insulated container. If it produces liquid water at 40°C [heat of vaporization of water is 540 cal/g and heat of fusion of ice is 80 cal/g], the value of M is _________.Correct answer is '40'. Can you explain this answer? has been provided alongside types of M grams of steam at 100°C is mixed with 200 g of ice at its melting point in a thermally insulated container. If it produces liquid water at 40°C [heat of vaporization of water is 540 cal/g and heat of fusion of ice is 80 cal/g], the value of M is _________.Correct answer is '40'. Can you explain this answer? theory, EduRev gives you an
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