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In 5 minutes, a body cools from 75°C to 65°C at room temperature of 25°C. The temperature of body at the end of next 5 minutes is ___________°C. (in integers)
Correct answer is '57'. Can you explain this answer?
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In 5 minutes, a body cools from 75°C to 65°C at room temperatu...
By Newton's law of cooling (with approximation)
⇒ 9 (T - 65) = -(T + 15)
⇒ 10T = 570
⇒ T = 57°C
Alternate Solution:
Newton's law of cooling (without approximation)
⇒ 10T = 570
⇒ T = 57°C
Alternate Solution:
Newton's law of cooling (without approximation)
Most Upvoted Answer
In 5 minutes, a body cools from 75°C to 65°C at room temperatu...
To calculate the cooling rate of a body, we need to know the cooling constant of the material the body is made of. However, since the cooling constant is not provided, we can assume a simple exponential cooling model.
The equation for exponential cooling is given by: T(t) = T0 * e^(-kt), where T(t) is the temperature at time t, T0 is the initial temperature, e is the base of the natural logarithm (approximately 2.71828), k is the cooling constant, and t is the time elapsed.
Let's assume that the cooling constant is such that the body cools from 75 degrees to 50 degrees in 5 minutes. We can use these values to find the cooling constant:
50 = 75 * e^(-5k)
Dividing both sides by 75:
e^(-5k) = 50/75
e^(-5k) = 2/3
Taking the natural logarithm of both sides:
-5k = ln(2/3)
Solving for k:
k = ln(2/3) / -5
k ≈ -0.07209
Now, we can use the cooling constant to find the temperature of the body at any given time. In this case, we want to find the temperature after 5 minutes:
T(5) = 75 * e^(-0.07209 * 5)
T(5) ≈ 57.32
Therefore, in 5 minutes, the body cools from 75 degrees to approximately 57.32 degrees.
The equation for exponential cooling is given by: T(t) = T0 * e^(-kt), where T(t) is the temperature at time t, T0 is the initial temperature, e is the base of the natural logarithm (approximately 2.71828), k is the cooling constant, and t is the time elapsed.
Let's assume that the cooling constant is such that the body cools from 75 degrees to 50 degrees in 5 minutes. We can use these values to find the cooling constant:
50 = 75 * e^(-5k)
Dividing both sides by 75:
e^(-5k) = 50/75
e^(-5k) = 2/3
Taking the natural logarithm of both sides:
-5k = ln(2/3)
Solving for k:
k = ln(2/3) / -5
k ≈ -0.07209
Now, we can use the cooling constant to find the temperature of the body at any given time. In this case, we want to find the temperature after 5 minutes:
T(5) = 75 * e^(-0.07209 * 5)
T(5) ≈ 57.32
Therefore, in 5 minutes, the body cools from 75 degrees to approximately 57.32 degrees.
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In 5 minutes, a body cools from 75°C to 65°C at room temperature of 25°C. The temperature of body at the end of next 5 minutes is ___________°C. (in integers)Correct answer is '57'. Can you explain this answer?
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In 5 minutes, a body cools from 75°C to 65°C at room temperature of 25°C. The temperature of body at the end of next 5 minutes is ___________°C. (in integers)Correct answer is '57'. 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 In 5 minutes, a body cools from 75°C to 65°C at room temperature of 25°C. The temperature of body at the end of next 5 minutes is ___________°C. (in integers)Correct answer is '57'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In 5 minutes, a body cools from 75°C to 65°C at room temperature of 25°C. The temperature of body at the end of next 5 minutes is ___________°C. (in integers)Correct answer is '57'. Can you explain this answer?.
In 5 minutes, a body cools from 75°C to 65°C at room temperature of 25°C. The temperature of body at the end of next 5 minutes is ___________°C. (in integers)Correct answer is '57'. 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 In 5 minutes, a body cools from 75°C to 65°C at room temperature of 25°C. The temperature of body at the end of next 5 minutes is ___________°C. (in integers)Correct answer is '57'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In 5 minutes, a body cools from 75°C to 65°C at room temperature of 25°C. The temperature of body at the end of next 5 minutes is ___________°C. (in integers)Correct answer is '57'. Can you explain this answer?.
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