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A body cools from 50.0°C to 48°C in 5s. How long will it take to cool from 40.0°C to 39°C?Assume the temperature of surroundings to be 30.0°C and Newton's law of cooling to be valid.
- a)2.5 sec
- b)10 sec
- c)20 sec
- d)9 sec
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A body cools from 50.0°C to 48°C in 5s. How long will it take ...
Rate of cooling ∝ temperature difference between system and surrounding.
As the temperature difference is halved, so the rate of cooling will also be halved
As the temperature difference is halved, so the rate of cooling will also be halved
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A body cools from 50.0°C to 48°C in 5s. How long will it take ...
°C to 20.0°C in 30 minutes. If the initial temperature of the surroundings is 20.0°C, what is the rate of heat loss from the body?
We can use the formula for rate of heat loss:
Q/t = kAΔT/L
where Q is the heat lost, t is the time, k is the thermal conductivity, A is the surface area, ΔT is the temperature difference, and L is the thickness of the material.
Assuming the body is a uniform solid and the heat loss is mainly due to conduction, we can simplify the formula as:
Q/t = kA(θi - θf)/d
where θi is the initial temperature, θf is the final temperature, and d is the distance between the two surfaces.
We can rearrange the formula to solve for kA/d, which represents the heat transfer coefficient:
kA/d = Q/t/(θi - θf)
Substituting the given values, we get:
kA/d = (mCp)(θi - θf)/t/(θi - θs)
where m is the mass of the body, Cp is the specific heat capacity, and θs is the temperature of the surroundings.
Assuming the body is made of copper, which has a thermal conductivity of k = 401 W/mK, and has a density of ρ = 8960 kg/m³ and a specific heat capacity of Cp = 385 J/kgK, we can calculate the surface area and thickness of the body:
A = m/ρ = (1000 g)/(8960 kg/m³) = 0.1118 m²
V = m/ρ = (1000 g)/(8960 kg/m³) = 0.0001118 m³
d = V/A = 0.0001118 m³/0.1118 m² = 0.001 m = 1 mm
Substituting the values, we get:
kA/d = (1000 g)(385 J/kgK)(50.0 - 20.0)°C/(30 min)(50.0 - 20.0)°C/(20.0)°C
kA/d = 9.63 W/m²K
Therefore, the rate of heat loss from the body is:
Q/t = kAΔT/L = (9.63 W/m²K)(0.1118 m²)(50.0 - 20.0)°C/0.001 m
Q/t = 1.87 × 10³ W or 1.87 kW
Answer: The rate of heat loss from the body is 1.87 kW.
We can use the formula for rate of heat loss:
Q/t = kAΔT/L
where Q is the heat lost, t is the time, k is the thermal conductivity, A is the surface area, ΔT is the temperature difference, and L is the thickness of the material.
Assuming the body is a uniform solid and the heat loss is mainly due to conduction, we can simplify the formula as:
Q/t = kA(θi - θf)/d
where θi is the initial temperature, θf is the final temperature, and d is the distance between the two surfaces.
We can rearrange the formula to solve for kA/d, which represents the heat transfer coefficient:
kA/d = Q/t/(θi - θf)
Substituting the given values, we get:
kA/d = (mCp)(θi - θf)/t/(θi - θs)
where m is the mass of the body, Cp is the specific heat capacity, and θs is the temperature of the surroundings.
Assuming the body is made of copper, which has a thermal conductivity of k = 401 W/mK, and has a density of ρ = 8960 kg/m³ and a specific heat capacity of Cp = 385 J/kgK, we can calculate the surface area and thickness of the body:
A = m/ρ = (1000 g)/(8960 kg/m³) = 0.1118 m²
V = m/ρ = (1000 g)/(8960 kg/m³) = 0.0001118 m³
d = V/A = 0.0001118 m³/0.1118 m² = 0.001 m = 1 mm
Substituting the values, we get:
kA/d = (1000 g)(385 J/kgK)(50.0 - 20.0)°C/(30 min)(50.0 - 20.0)°C/(20.0)°C
kA/d = 9.63 W/m²K
Therefore, the rate of heat loss from the body is:
Q/t = kAΔT/L = (9.63 W/m²K)(0.1118 m²)(50.0 - 20.0)°C/0.001 m
Q/t = 1.87 × 10³ W or 1.87 kW
Answer: The rate of heat loss from the body is 1.87 kW.
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A body cools from 50.0°C to 48°C in 5s. How long will it take ...
Ans will be 5s not 10s. Apply newtons law of cooling formula 5s is the ans
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A body cools from 50.0°C to 48°C in 5s. How long will it take to cool from 40.0°C to 39°C?Assume the temperature of surroundings to be 30.0°C and Newtons law of cooling to be valid.a)2.5 secb)10 secc)20 secd)9 secCorrect answer is option 'B'. Can you explain this answer?
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A body cools from 50.0°C to 48°C in 5s. How long will it take to cool from 40.0°C to 39°C?Assume the temperature of surroundings to be 30.0°C and Newtons law of cooling to be valid.a)2.5 secb)10 secc)20 secd)9 secCorrect answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A body cools from 50.0°C to 48°C in 5s. How long will it take to cool from 40.0°C to 39°C?Assume the temperature of surroundings to be 30.0°C and Newtons law of cooling to be valid.a)2.5 secb)10 secc)20 secd)9 secCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A body cools from 50.0°C to 48°C in 5s. How long will it take to cool from 40.0°C to 39°C?Assume the temperature of surroundings to be 30.0°C and Newtons law of cooling to be valid.a)2.5 secb)10 secc)20 secd)9 secCorrect answer is option 'B'. Can you explain this answer?.
A body cools from 50.0°C to 48°C in 5s. How long will it take to cool from 40.0°C to 39°C?Assume the temperature of surroundings to be 30.0°C and Newtons law of cooling to be valid.a)2.5 secb)10 secc)20 secd)9 secCorrect answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A body cools from 50.0°C to 48°C in 5s. How long will it take to cool from 40.0°C to 39°C?Assume the temperature of surroundings to be 30.0°C and Newtons law of cooling to be valid.a)2.5 secb)10 secc)20 secd)9 secCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A body cools from 50.0°C to 48°C in 5s. How long will it take to cool from 40.0°C to 39°C?Assume the temperature of surroundings to be 30.0°C and Newtons law of cooling to be valid.a)2.5 secb)10 secc)20 secd)9 secCorrect answer is option 'B'. Can you explain this answer?.
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