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An ice cube of mass 0.1 kg at 0°C is placed in an isolated container which is at 227° C. The specific heat S of the container varies with temperature T according to the container to the empirical relation S = A + BT, where A = 100 cal/kg.k and B = 2 x 10 2 cal/kg.k2. If the final temperature of the container is 27° C. Then the mass of the container is ______ kg.
[Latent heat of fusion for water = 8 x 104 cal/kg. specific heat of water = 103 cal/ kg.k ]
Give answer upto to 1decimal
[Latent heat of fusion for water = 8 x 104 cal/kg. specific heat of water = 103 cal/ kg.k ]
Give answer upto to 1decimal
Correct answer is '0.5'. Can you explain this answer?
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Verified Answer
An ice cube of mass 0.1 kg at 0°C is placed in an isolated contain...
0.49 or 0.5
Let m be the mass of the container Initial temperature of container
T1 = (227 + 273) = 500kFinal temperature of container TF = 27 + 273 = 300 k
Now Heat gained by the ice cube = Heat lost by the container.
⇒ (mass of ice) (Latent heat of fusion of ice) + (mass of ice) (specific heat of water) (300-273)
⇒ (0.1)(8x104)+ (0.1)(103)27
(A + BT)dT
10700 =
m = 0.495 kg
m=0.5 kg
Let m be the mass of the container Initial temperature of container
T1 = (227 + 273) = 500kFinal temperature of container TF = 27 + 273 = 300 k
Now Heat gained by the ice cube = Heat lost by the container.
⇒ (mass of ice) (Latent heat of fusion of ice) + (mass of ice) (specific heat of water) (300-273)
⇒ (0.1)(8x104)+ (0.1)(103)27
(A + BT)dT10700 =
m = 0.495 kg
m=0.5 kg
Most Upvoted Answer
An ice cube of mass 0.1 kg at 0°C is placed in an isolated contain...
To find the final temperature of the ice cube, we can use the equation:
Q = mcΔT
Where:
Q = heat transferred
m = mass of the ice cube
c = specific heat capacity of ice
ΔT = change in temperature
First, let's calculate the heat transferred. Since the ice cube is initially at 0°C and we want to find the final temperature, we can assume it reaches the melting point of ice, which is 0°C. Therefore, ΔT = 0°C - 0°C = 0°C.
Next, we need to find the specific heat capacity of ice. The specific heat capacity of ice is 2,093 J/kg°C.
Now we can calculate the heat transferred:
Q = (0.1 kg)(2,093 J/kg°C)(0°C - 0°C)
Q = 0 J
Since no heat is transferred, the final temperature of the ice cube remains at 0°C.
Q = mcΔT
Where:
Q = heat transferred
m = mass of the ice cube
c = specific heat capacity of ice
ΔT = change in temperature
First, let's calculate the heat transferred. Since the ice cube is initially at 0°C and we want to find the final temperature, we can assume it reaches the melting point of ice, which is 0°C. Therefore, ΔT = 0°C - 0°C = 0°C.
Next, we need to find the specific heat capacity of ice. The specific heat capacity of ice is 2,093 J/kg°C.
Now we can calculate the heat transferred:
Q = (0.1 kg)(2,093 J/kg°C)(0°C - 0°C)
Q = 0 J
Since no heat is transferred, the final temperature of the ice cube remains at 0°C.
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An ice cube of mass 0.1 kg at 0°C is placed in an isolated container which is at 227° C. The specific heat S of the container varies with temperature T according to the container to the empirical relation S = A + BT, where A = 100 cal/kg.k and B = 2 x 10 2 cal/kg.k2. If the final temperature of the container is 27° C. Thenthe mass of the container is ______ kg.[Latent heat of fusion for water = 8 x 104 cal/kg. specific heat of water = 103 cal/kg.k ]Give answer upto to 1decimalCorrect answer is '0.5'. Can you explain this answer?
Question Description
An ice cube of mass 0.1 kg at 0°C is placed in an isolated container which is at 227° C. The specific heat S of the container varies with temperature T according to the container to the empirical relation S = A + BT, where A = 100 cal/kg.k and B = 2 x 10 2 cal/kg.k2. If the final temperature of the container is 27° C. Thenthe mass of the container is ______ kg.[Latent heat of fusion for water = 8 x 104 cal/kg. specific heat of water = 103 cal/kg.k ]Give answer upto to 1decimalCorrect answer is '0.5'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about An ice cube of mass 0.1 kg at 0°C is placed in an isolated container which is at 227° C. The specific heat S of the container varies with temperature T according to the container to the empirical relation S = A + BT, where A = 100 cal/kg.k and B = 2 x 10 2 cal/kg.k2. If the final temperature of the container is 27° C. Thenthe mass of the container is ______ kg.[Latent heat of fusion for water = 8 x 104 cal/kg. specific heat of water = 103 cal/kg.k ]Give answer upto to 1decimalCorrect answer is '0.5'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An ice cube of mass 0.1 kg at 0°C is placed in an isolated container which is at 227° C. The specific heat S of the container varies with temperature T according to the container to the empirical relation S = A + BT, where A = 100 cal/kg.k and B = 2 x 10 2 cal/kg.k2. If the final temperature of the container is 27° C. Thenthe mass of the container is ______ kg.[Latent heat of fusion for water = 8 x 104 cal/kg. specific heat of water = 103 cal/kg.k ]Give answer upto to 1decimalCorrect answer is '0.5'. Can you explain this answer?.
An ice cube of mass 0.1 kg at 0°C is placed in an isolated container which is at 227° C. The specific heat S of the container varies with temperature T according to the container to the empirical relation S = A + BT, where A = 100 cal/kg.k and B = 2 x 10 2 cal/kg.k2. If the final temperature of the container is 27° C. Thenthe mass of the container is ______ kg.[Latent heat of fusion for water = 8 x 104 cal/kg. specific heat of water = 103 cal/kg.k ]Give answer upto to 1decimalCorrect answer is '0.5'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about An ice cube of mass 0.1 kg at 0°C is placed in an isolated container which is at 227° C. The specific heat S of the container varies with temperature T according to the container to the empirical relation S = A + BT, where A = 100 cal/kg.k and B = 2 x 10 2 cal/kg.k2. If the final temperature of the container is 27° C. Thenthe mass of the container is ______ kg.[Latent heat of fusion for water = 8 x 104 cal/kg. specific heat of water = 103 cal/kg.k ]Give answer upto to 1decimalCorrect answer is '0.5'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An ice cube of mass 0.1 kg at 0°C is placed in an isolated container which is at 227° C. The specific heat S of the container varies with temperature T according to the container to the empirical relation S = A + BT, where A = 100 cal/kg.k and B = 2 x 10 2 cal/kg.k2. If the final temperature of the container is 27° C. Thenthe mass of the container is ______ kg.[Latent heat of fusion for water = 8 x 104 cal/kg. specific heat of water = 103 cal/kg.k ]Give answer upto to 1decimalCorrect answer is '0.5'. Can you explain this answer?.
Solutions for An ice cube of mass 0.1 kg at 0°C is placed in an isolated container which is at 227° C. The specific heat S of the container varies with temperature T according to the container to the empirical relation S = A + BT, where A = 100 cal/kg.k and B = 2 x 10 2 cal/kg.k2. If the final temperature of the container is 27° C. Thenthe mass of the container is ______ kg.[Latent heat of fusion for water = 8 x 104 cal/kg. specific heat of water = 103 cal/kg.k ]Give answer upto to 1decimalCorrect answer is '0.5'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE.
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An ice cube of mass 0.1 kg at 0°C is placed in an isolated container which is at 227° C. The specific heat S of the container varies with temperature T according to the container to the empirical relation S = A + BT, where A = 100 cal/kg.k and B = 2 x 10 2 cal/kg.k2. If the final temperature of the container is 27° C. Thenthe mass of the container is ______ kg.[Latent heat of fusion for water = 8 x 104 cal/kg. specific heat of water = 103 cal/kg.k ]Give answer upto to 1decimalCorrect answer is '0.5'. Can you explain this answer?, a detailed solution for An ice cube of mass 0.1 kg at 0°C is placed in an isolated container which is at 227° C. The specific heat S of the container varies with temperature T according to the container to the empirical relation S = A + BT, where A = 100 cal/kg.k and B = 2 x 10 2 cal/kg.k2. If the final temperature of the container is 27° C. Thenthe mass of the container is ______ kg.[Latent heat of fusion for water = 8 x 104 cal/kg. specific heat of water = 103 cal/kg.k ]Give answer upto to 1decimalCorrect answer is '0.5'. Can you explain this answer? has been provided alongside types of An ice cube of mass 0.1 kg at 0°C is placed in an isolated container which is at 227° C. The specific heat S of the container varies with temperature T according to the container to the empirical relation S = A + BT, where A = 100 cal/kg.k and B = 2 x 10 2 cal/kg.k2. If the final temperature of the container is 27° C. Thenthe mass of the container is ______ kg.[Latent heat of fusion for water = 8 x 104 cal/kg. specific heat of water = 103 cal/kg.k ]Give answer upto to 1decimalCorrect answer is '0.5'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice An ice cube of mass 0.1 kg at 0°C is placed in an isolated container which is at 227° C. The specific heat S of the container varies with temperature T according to the container to the empirical relation S = A + BT, where A = 100 cal/kg.k and B = 2 x 10 2 cal/kg.k2. If the final temperature of the container is 27° C. Thenthe mass of the container is ______ kg.[Latent heat of fusion for water = 8 x 104 cal/kg. specific heat of water = 103 cal/kg.k ]Give answer upto to 1decimalCorrect answer is '0.5'. Can you explain this answer? tests, examples and also practice GATE tests.
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