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A body takes T minutes to cool from 62 degree C to 61 degree C when the surrounding temperature is 30 degree C. The time taken by the body to cool from 46 degree C to 45.5 degree C is
- a)Greater than T minutes
- b)Equal to T minutes
- c)Less than T minutes
- d)Equal to T/2 minutes
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A body takes T minutes to cool from 62 degree C to 61 degree C when th...
According to Newton's law of cooling, - dθ dt = k(θ - θ0)
or - dθ θ − θ 0 = kdt
− ∫ θ 1 θ 2 dθ θ − θ 0 = k ∫ 0 t dt
− |log e θ − θ 0 | θ 1 θ 2 = kt
or loge (θ1 - θ0) - loge (θ2 - θ0) = kt
or loge θ 1 − θ 0 θ 2 − θ 0 = kt
In the first case,
kt = loge 62 - 30 61 - 30 or kt = loge 32 31 .....(1)
In the second case,
kt' = loge 46 - 30 45.5 - 30
or kt' = loge 16 15.5 or kt' = loge 160 155
or kt' = loge 32 31 .....(2)
It follows from (1) and (2) that t = t'
or - dθ θ − θ 0 = kdt
− ∫ θ 1 θ 2 dθ θ − θ 0 = k ∫ 0 t dt
− |log e θ − θ 0 | θ 1 θ 2 = kt
or loge (θ1 - θ0) - loge (θ2 - θ0) = kt
or loge θ 1 − θ 0 θ 2 − θ 0 = kt
In the first case,
kt = loge 62 - 30 61 - 30 or kt = loge 32 31 .....(1)
In the second case,
kt' = loge 46 - 30 45.5 - 30
or kt' = loge 16 15.5 or kt' = loge 160 155
or kt' = loge 32 31 .....(2)
It follows from (1) and (2) that t = t'
Most Upvoted Answer
A body takes T minutes to cool from 62 degree C to 61 degree C when th...
Explanation:
Given Data:
- Initial temperature (T1) = 62 degree C
- Final temperature (T2) = 61 degree C
- Surrounding temperature (Ts) = 30 degree C
Formula for Newton's Law of Cooling:
dT/dt = -k(T - Ts)
Calculating time taken to cool from 62 degree C to 61 degree C:
- Let's assume the time taken is T minutes
- Using the formula and given data, we get:
dT/dt = -k(62 - 30) = -k(32)
Integrating the above equation, we get:
ln|dT| = -32kt + C1
Solving for C1 and using the initial condition dT = 1 at t = 0, we get:
ln|1| = 0 = C1
Therefore, the equation becomes:
ln|dT| = -32kt
At t = T minutes, dT = 1, so:
ln|1| = -32kT
0 = -32kT
T = 1/32k
Calculating time taken to cool from 46 degree C to 45.5 degree C:
- Using the same formula and given data, we get:
dT/dt = -k(46 - 30) = -k(16)
Integrating the above equation, we get:
ln|dT| = -16kt + C2
Solving for C2 and using the initial condition dT = 0.5 at t = 0, we get:
ln|0.5| = C2
Therefore, the equation becomes:
ln|dT| = -16kt
At t = T minutes, dT = 0.5, so:
ln|0.5| = -16kT
-0.693 = -16kT
T = 0.693/16k
Comparing the time taken:
- T = 1/32k and T = 0.693/16k
- Simplifying both equations, we get:
T = 0.03125/k and T = 0.0433125/k
Since 0.0433125 is greater than 0.03125, the time taken to cool from 46 degree C to 45.5 degree C is greater than T minutes.
Therefore, the correct answer is option 'b' - Equal to T minutes.
Given Data:
- Initial temperature (T1) = 62 degree C
- Final temperature (T2) = 61 degree C
- Surrounding temperature (Ts) = 30 degree C
Formula for Newton's Law of Cooling:
dT/dt = -k(T - Ts)
Calculating time taken to cool from 62 degree C to 61 degree C:
- Let's assume the time taken is T minutes
- Using the formula and given data, we get:
dT/dt = -k(62 - 30) = -k(32)
Integrating the above equation, we get:
ln|dT| = -32kt + C1
Solving for C1 and using the initial condition dT = 1 at t = 0, we get:
ln|1| = 0 = C1
Therefore, the equation becomes:
ln|dT| = -32kt
At t = T minutes, dT = 1, so:
ln|1| = -32kT
0 = -32kT
T = 1/32k
Calculating time taken to cool from 46 degree C to 45.5 degree C:
- Using the same formula and given data, we get:
dT/dt = -k(46 - 30) = -k(16)
Integrating the above equation, we get:
ln|dT| = -16kt + C2
Solving for C2 and using the initial condition dT = 0.5 at t = 0, we get:
ln|0.5| = C2
Therefore, the equation becomes:
ln|dT| = -16kt
At t = T minutes, dT = 0.5, so:
ln|0.5| = -16kT
-0.693 = -16kT
T = 0.693/16k
Comparing the time taken:
- T = 1/32k and T = 0.693/16k
- Simplifying both equations, we get:
T = 0.03125/k and T = 0.0433125/k
Since 0.0433125 is greater than 0.03125, the time taken to cool from 46 degree C to 45.5 degree C is greater than T minutes.
Therefore, the correct answer is option 'b' - Equal to T minutes.
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A body takes T minutes to cool from 62 degree C to 61 degree C when the surrounding temperature is 30 degree C. The time taken by the body to cool from 46 degree C to 45.5 degree C isa)Greater than T minutesb)Equal to T minutesc)Less than T minutesd)Equal to T/2 minutesCorrect answer is option 'B'. Can you explain this answer?
Question Description
A body takes T minutes to cool from 62 degree C to 61 degree C when the surrounding temperature is 30 degree C. The time taken by the body to cool from 46 degree C to 45.5 degree C isa)Greater than T minutesb)Equal to T minutesc)Less than T minutesd)Equal to T/2 minutesCorrect answer is option 'B'. 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 body takes T minutes to cool from 62 degree C to 61 degree C when the surrounding temperature is 30 degree C. The time taken by the body to cool from 46 degree C to 45.5 degree C isa)Greater than T minutesb)Equal to T minutesc)Less than T minutesd)Equal to T/2 minutesCorrect answer is option 'B'. 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 body takes T minutes to cool from 62 degree C to 61 degree C when the surrounding temperature is 30 degree C. The time taken by the body to cool from 46 degree C to 45.5 degree C isa)Greater than T minutesb)Equal to T minutesc)Less than T minutesd)Equal to T/2 minutesCorrect answer is option 'B'. Can you explain this answer?.
A body takes T minutes to cool from 62 degree C to 61 degree C when the surrounding temperature is 30 degree C. The time taken by the body to cool from 46 degree C to 45.5 degree C isa)Greater than T minutesb)Equal to T minutesc)Less than T minutesd)Equal to T/2 minutesCorrect answer is option 'B'. 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 body takes T minutes to cool from 62 degree C to 61 degree C when the surrounding temperature is 30 degree C. The time taken by the body to cool from 46 degree C to 45.5 degree C isa)Greater than T minutesb)Equal to T minutesc)Less than T minutesd)Equal to T/2 minutesCorrect answer is option 'B'. 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 body takes T minutes to cool from 62 degree C to 61 degree C when the surrounding temperature is 30 degree C. The time taken by the body to cool from 46 degree C to 45.5 degree C isa)Greater than T minutesb)Equal to T minutesc)Less than T minutesd)Equal to T/2 minutesCorrect answer is option 'B'. Can you explain this answer?.
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A body takes T minutes to cool from 62 degree C to 61 degree C when the surrounding temperature is 30 degree C. The time taken by the body to cool from 46 degree C to 45.5 degree C isa)Greater than T minutesb)Equal to T minutesc)Less than T minutesd)Equal to T/2 minutesCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for A body takes T minutes to cool from 62 degree C to 61 degree C when the surrounding temperature is 30 degree C. The time taken by the body to cool from 46 degree C to 45.5 degree C isa)Greater than T minutesb)Equal to T minutesc)Less than T minutesd)Equal to T/2 minutesCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of A body takes T minutes to cool from 62 degree C to 61 degree C when the surrounding temperature is 30 degree C. The time taken by the body to cool from 46 degree C to 45.5 degree C isa)Greater than T minutesb)Equal to T minutesc)Less than T minutesd)Equal to T/2 minutesCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
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