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A body cools from 50 degree c to 40 degree c in 5 minutes . Its temperature comes down to 33.33 degree c in the next 5 minutes . The temperature of the surrounding is 1) 15 degree c 2) 20 degree c 3) 25 degree c 4) 10 degree c?
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A body cools from 50 degree c to 40 degree c in 5 minutes . Its temper...
Given:
Initial temperature of the body, T1 = 50°C
Final temperature of the body, T2 = 33.33°C
Time taken to cool from T1 to T2, t1 = 5 minutes
Time taken to cool from T2 to Tsurrounding, t2 = 5 minutes
To find: Temperature of the surrounding, Tsurrounding
Formula: Newton's Law of Cooling
dT/dt = -k(T - Tsurrounding)
Where,
dT/dt: Rate of change of temperature of the body with respect to time
k: Constant of proportionality
T: Temperature of the body at any time
Tsurrounding: Temperature of the surrounding
Calculation:
Step 1: Finding the constant of proportionality, k
dT/dt = -k(T - Tsurrounding)
From the given data,
When T = 50°C, dT/dt = -k(50 - Tsurrounding)
When T = 40°C, dT/dt = -k(40 - Tsurrounding)
Using these two equations, we can find k
(dT/dt)1 / (dT/dt)2 = (50 - Tsurrounding)/(40 - Tsurrounding)
(-16.67)/(-12.5) = (50 - Tsurrounding)/(40 - Tsurrounding)
(50 - Tsurrounding)/(40 - Tsurrounding) = 4/3
150 - 3Tsurrounding = 160 - 4Tsurrounding
Tsurrounding = 10°C
Step 2: Verifying the answer
dT/dt = -k(T - Tsurrounding)
When T = 33.33°C, dT/dt = -k(33.33 - 10) = -k(23.33)
When t = 5 minutes, dT/dt = (T2 - Tsurrounding)/t2 = (33.33 - 10)/5 = 4.6667
So, -k(23.33) = 4.6667
k = -4.6667/-23.33 = 0.2
dT/dt = -k(T - Tsurrounding)
When T = 33.33°C, dT/dt = -0.2(33.33 - 10) = -4.6667
This matches with the value we found earlier, so our answer is correct.
Therefore, the temperature of the surrounding is 10°C.
Answer: 4) 10 degree c
Initial temperature of the body, T1 = 50°C
Final temperature of the body, T2 = 33.33°C
Time taken to cool from T1 to T2, t1 = 5 minutes
Time taken to cool from T2 to Tsurrounding, t2 = 5 minutes
To find: Temperature of the surrounding, Tsurrounding
Formula: Newton's Law of Cooling
dT/dt = -k(T - Tsurrounding)
Where,
dT/dt: Rate of change of temperature of the body with respect to time
k: Constant of proportionality
T: Temperature of the body at any time
Tsurrounding: Temperature of the surrounding
Calculation:
Step 1: Finding the constant of proportionality, k
dT/dt = -k(T - Tsurrounding)
From the given data,
When T = 50°C, dT/dt = -k(50 - Tsurrounding)
When T = 40°C, dT/dt = -k(40 - Tsurrounding)
Using these two equations, we can find k
(dT/dt)1 / (dT/dt)2 = (50 - Tsurrounding)/(40 - Tsurrounding)
(-16.67)/(-12.5) = (50 - Tsurrounding)/(40 - Tsurrounding)
(50 - Tsurrounding)/(40 - Tsurrounding) = 4/3
150 - 3Tsurrounding = 160 - 4Tsurrounding
Tsurrounding = 10°C
Step 2: Verifying the answer
dT/dt = -k(T - Tsurrounding)
When T = 33.33°C, dT/dt = -k(33.33 - 10) = -k(23.33)
When t = 5 minutes, dT/dt = (T2 - Tsurrounding)/t2 = (33.33 - 10)/5 = 4.6667
So, -k(23.33) = 4.6667
k = -4.6667/-23.33 = 0.2
dT/dt = -k(T - Tsurrounding)
When T = 33.33°C, dT/dt = -0.2(33.33 - 10) = -4.6667
This matches with the value we found earlier, so our answer is correct.
Therefore, the temperature of the surrounding is 10°C.
Answer: 4) 10 degree c
Community Answer
A body cools from 50 degree c to 40 degree c in 5 minutes . Its temper...
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A body cools from 50 degree c to 40 degree c in 5 minutes . Its temperature comes down to 33.33 degree c in the next 5 minutes . The temperature of the surrounding is 1) 15 degree c 2) 20 degree c 3) 25 degree c 4) 10 degree c?
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A body cools from 50 degree c to 40 degree c in 5 minutes . Its temperature comes down to 33.33 degree c in the next 5 minutes . The temperature of the surrounding is 1) 15 degree c 2) 20 degree c 3) 25 degree c 4) 10 degree c? 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 degree c to 40 degree c in 5 minutes . Its temperature comes down to 33.33 degree c in the next 5 minutes . The temperature of the surrounding is 1) 15 degree c 2) 20 degree c 3) 25 degree c 4) 10 degree c? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A body cools from 50 degree c to 40 degree c in 5 minutes . Its temperature comes down to 33.33 degree c in the next 5 minutes . The temperature of the surrounding is 1) 15 degree c 2) 20 degree c 3) 25 degree c 4) 10 degree c?.
A body cools from 50 degree c to 40 degree c in 5 minutes . Its temperature comes down to 33.33 degree c in the next 5 minutes . The temperature of the surrounding is 1) 15 degree c 2) 20 degree c 3) 25 degree c 4) 10 degree c? 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 degree c to 40 degree c in 5 minutes . Its temperature comes down to 33.33 degree c in the next 5 minutes . The temperature of the surrounding is 1) 15 degree c 2) 20 degree c 3) 25 degree c 4) 10 degree c? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A body cools from 50 degree c to 40 degree c in 5 minutes . Its temperature comes down to 33.33 degree c in the next 5 minutes . The temperature of the surrounding is 1) 15 degree c 2) 20 degree c 3) 25 degree c 4) 10 degree c?.
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