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A block of ice of mass 120 g at temperature 0°C is put in 300 gm of water at 25°C. The x g of ice melts as the temperature of the water reaches 0°C. The value of x is (in integers)
[Use: Specific heat capacity of water = 4,200 Jkg-1K-1, Latent heat of ice = 3.5 × 105 Jkg-1]
[Use: Specific heat capacity of water = 4,200 Jkg-1K-1, Latent heat of ice = 3.5 × 105 Jkg-1]
Correct answer is '90'. Can you explain this answer?
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A block of ice of mass 120 g at temperature 0°C is put in 300 gm o...
°C is placed in a cup of hot water at temperature 80°C. Assuming no heat is lost to the surroundings, calculate the final temperature of the mixture when thermal equilibrium is reached. (Specific heat capacity of water = 4.18 J/g°C, specific heat capacity of ice = 2.09 J/g°C, heat of fusion of ice = 333 J/g)
To solve this problem, we need to calculate the heat gained by the ice and the heat lost by the hot water, and then equate them to find the final temperature.
First, let's calculate the heat gained by the ice using the equation:
Q_ice = m_ice * c_ice * ΔT_ice
where Q_ice is the heat gained by the ice, m_ice is the mass of the ice (120 g), c_ice is the specific heat capacity of ice (2.09 J/g°C), and ΔT_ice is the change in temperature of the ice.
ΔT_ice = Tf - Ti
where Tf is the final temperature of the mixture and Ti is the initial temperature of the ice (0°C).
ΔT_ice = Tf - 0 = Tf
Q_ice = 120 g * 2.09 J/g°C * Tf = 250.8 Tf J
Next, let's calculate the heat lost by the hot water using the equation:
Q_water = m_water * c_water * ΔT_water
where Q_water is the heat lost by the hot water, m_water is the mass of the hot water, c_water is the specific heat capacity of water (4.18 J/g°C), and ΔT_water is the change in temperature of the hot water.
ΔT_water = Tf - Ti
where Tf is the final temperature of the mixture and Ti is the initial temperature of the hot water (80°C).
ΔT_water = Tf - 80
Q_water = m_water * 4.18 J/g°C * (Tf - 80) = 4.18 m_water (Tf - 80) J
Since the heat gained by the ice equals the heat lost by the hot water, we can equate the two equations:
250.8 Tf = 4.18 m_water (Tf - 80)
Now, we can substitute the mass of the water into the equation:
m_water = Q_ice / (c_water * (Tf - 80))
m_water = 250.8 Tf / (4.18 * (Tf - 80))
To find the final temperature Tf, we need to solve the equation:
250.8 Tf = 4.18 * 120 / (4.18 * (Tf - 80))
250.8 Tf = 120 / (Tf - 80)
250.8 Tf * (Tf - 80) = 120
250.8 Tf^2 - 20160 Tf + 1929600 = 120
250.8 Tf^2 - 20160 Tf + 1929480 = 0
Using the quadratic formula, we can solve for Tf:
Tf = (-(-20160) ± √((-20160)^2 - 4 * 250.8 * 1929480)) / (2 * 250.8)
Tf = (20160 ± √(406425600
To solve this problem, we need to calculate the heat gained by the ice and the heat lost by the hot water, and then equate them to find the final temperature.
First, let's calculate the heat gained by the ice using the equation:
Q_ice = m_ice * c_ice * ΔT_ice
where Q_ice is the heat gained by the ice, m_ice is the mass of the ice (120 g), c_ice is the specific heat capacity of ice (2.09 J/g°C), and ΔT_ice is the change in temperature of the ice.
ΔT_ice = Tf - Ti
where Tf is the final temperature of the mixture and Ti is the initial temperature of the ice (0°C).
ΔT_ice = Tf - 0 = Tf
Q_ice = 120 g * 2.09 J/g°C * Tf = 250.8 Tf J
Next, let's calculate the heat lost by the hot water using the equation:
Q_water = m_water * c_water * ΔT_water
where Q_water is the heat lost by the hot water, m_water is the mass of the hot water, c_water is the specific heat capacity of water (4.18 J/g°C), and ΔT_water is the change in temperature of the hot water.
ΔT_water = Tf - Ti
where Tf is the final temperature of the mixture and Ti is the initial temperature of the hot water (80°C).
ΔT_water = Tf - 80
Q_water = m_water * 4.18 J/g°C * (Tf - 80) = 4.18 m_water (Tf - 80) J
Since the heat gained by the ice equals the heat lost by the hot water, we can equate the two equations:
250.8 Tf = 4.18 m_water (Tf - 80)
Now, we can substitute the mass of the water into the equation:
m_water = Q_ice / (c_water * (Tf - 80))
m_water = 250.8 Tf / (4.18 * (Tf - 80))
To find the final temperature Tf, we need to solve the equation:
250.8 Tf = 4.18 * 120 / (4.18 * (Tf - 80))
250.8 Tf = 120 / (Tf - 80)
250.8 Tf * (Tf - 80) = 120
250.8 Tf^2 - 20160 Tf + 1929600 = 120
250.8 Tf^2 - 20160 Tf + 1929480 = 0
Using the quadratic formula, we can solve for Tf:
Tf = (-(-20160) ± √((-20160)^2 - 4 * 250.8 * 1929480)) / (2 * 250.8)
Tf = (20160 ± √(406425600
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Community Answer
A block of ice of mass 120 g at temperature 0°C is put in 300 gm o...
Energy released by water = 0.3 × 25 × 4,200 = 31,500 J
Suppose m kg of ice melts.
m × 3.5 × 105 = 31,500
m = 0.09 kg = 90 gm
x = 90
Suppose m kg of ice melts.
m × 3.5 × 105 = 31,500
m = 0.09 kg = 90 gm
x = 90
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A block of ice of mass 120 g at temperature 0°C is put in 300 gm of water at 25°C. The x g of ice melts as the temperature of the water reaches 0°C. The value of x is (in integers)[Use: Specific heat capacity of water = 4,200 Jkg-1K-1, Latent heat of ice = 3.5 × 105Jkg-1]Correct answer is '90'. Can you explain this answer?
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A block of ice of mass 120 g at temperature 0°C is put in 300 gm of water at 25°C. The x g of ice melts as the temperature of the water reaches 0°C. The value of x is (in integers)[Use: Specific heat capacity of water = 4,200 Jkg-1K-1, Latent heat of ice = 3.5 × 105Jkg-1]Correct answer is '90'. 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 block of ice of mass 120 g at temperature 0°C is put in 300 gm of water at 25°C. The x g of ice melts as the temperature of the water reaches 0°C. The value of x is (in integers)[Use: Specific heat capacity of water = 4,200 Jkg-1K-1, Latent heat of ice = 3.5 × 105Jkg-1]Correct answer is '90'. 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 block of ice of mass 120 g at temperature 0°C is put in 300 gm of water at 25°C. The x g of ice melts as the temperature of the water reaches 0°C. The value of x is (in integers)[Use: Specific heat capacity of water = 4,200 Jkg-1K-1, Latent heat of ice = 3.5 × 105Jkg-1]Correct answer is '90'. Can you explain this answer?.
A block of ice of mass 120 g at temperature 0°C is put in 300 gm of water at 25°C. The x g of ice melts as the temperature of the water reaches 0°C. The value of x is (in integers)[Use: Specific heat capacity of water = 4,200 Jkg-1K-1, Latent heat of ice = 3.5 × 105Jkg-1]Correct answer is '90'. 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 block of ice of mass 120 g at temperature 0°C is put in 300 gm of water at 25°C. The x g of ice melts as the temperature of the water reaches 0°C. The value of x is (in integers)[Use: Specific heat capacity of water = 4,200 Jkg-1K-1, Latent heat of ice = 3.5 × 105Jkg-1]Correct answer is '90'. 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 block of ice of mass 120 g at temperature 0°C is put in 300 gm of water at 25°C. The x g of ice melts as the temperature of the water reaches 0°C. The value of x is (in integers)[Use: Specific heat capacity of water = 4,200 Jkg-1K-1, Latent heat of ice = 3.5 × 105Jkg-1]Correct answer is '90'. Can you explain this answer?.
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A block of ice of mass 120 g at temperature 0°C is put in 300 gm of water at 25°C. The x g of ice melts as the temperature of the water reaches 0°C. The value of x is (in integers)[Use: Specific heat capacity of water = 4,200 Jkg-1K-1, Latent heat of ice = 3.5 × 105Jkg-1]Correct answer is '90'. Can you explain this answer?, a detailed solution for A block of ice of mass 120 g at temperature 0°C is put in 300 gm of water at 25°C. The x g of ice melts as the temperature of the water reaches 0°C. The value of x is (in integers)[Use: Specific heat capacity of water = 4,200 Jkg-1K-1, Latent heat of ice = 3.5 × 105Jkg-1]Correct answer is '90'. Can you explain this answer? has been provided alongside types of A block of ice of mass 120 g at temperature 0°C is put in 300 gm of water at 25°C. The x g of ice melts as the temperature of the water reaches 0°C. The value of x is (in integers)[Use: Specific heat capacity of water = 4,200 Jkg-1K-1, Latent heat of ice = 3.5 × 105Jkg-1]Correct answer is '90'. Can you explain this answer? theory, EduRev gives you an
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