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In a room where the temprature is 30°C, a body cods from 61°C to 59°C in 4 minutes. The time (in min.) taken by the body to cool from 51°C to 49°C will be
- a)4
- b)6
- c)5
- d)8
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
In a room where the temprature is 30°C, a body cods from 61°C ...
A.c. ..to Newton's law of cooling,
dx/dt=k(x-x0)..
so,
1st find d value of k from 1st condition then keep it in 2nd condition n find d time..
61-59/4=k((61+59)/2 -30)
1/2=k(60-30)
..it means k =1/60..
now in 2nd case we have k..n we have to find time i.e...dt..
51-49/dt=1/60×{(51+49)/2-30}
2/dt=1/60×{20}
..this will give you dt..as 6..i.e..time is 6 minutes...
dx/dt=k(x-x0)..
so,
1st find d value of k from 1st condition then keep it in 2nd condition n find d time..
61-59/4=k((61+59)/2 -30)
1/2=k(60-30)
..it means k =1/60..
now in 2nd case we have k..n we have to find time i.e...dt..
51-49/dt=1/60×{(51+49)/2-30}
2/dt=1/60×{20}
..this will give you dt..as 6..i.e..time is 6 minutes...
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In a room where the temprature is 30°C, a body cods from 61°C ...
Given Information:
Temperature in the room: 30°C
Initial temperature of the body: 61°C
Final temperature of the body: 59°C
Time taken to cool from 61°C to 59°C: 4 minutes
Explanation:
The rate at which a body cools is directly proportional to the temperature difference between the body and its surroundings. This is given by Newton's Law of Cooling.
Using the formula:
θ(t) = θ_s + (θ_i - θ_s)e^(-kt)
Where:
θ(t) = temperature at time t
θ_s = temperature of surroundings
θ_i = initial temperature of the body
k = cooling constant
t = time
Calculating the cooling constant:
Using the given data, we can find the cooling constant:
59 = 30 + (61 - 30)e^(-4k)
29 = 31e^(-4k)
e^(-4k) = 29/31
-4k = ln(29/31)
k = - ln(29/31) / 4
Finding the time taken to cool from 51°C to 49°C:
Using the formula with the new temperatures:
49 = 30 + (51 - 30)e^(-k*t)
19 = 21e^(-k*t)
e^(-k*t) = 19/21
t = -ln(19/21) / k
Substitute the value of k into the above equation to find t.
After solving, t ≈ 6 minutes.
Therefore, the time taken by the body to cool from 51°C to 49°C will be 6 minutes. Option (b) is correct.
Temperature in the room: 30°C
Initial temperature of the body: 61°C
Final temperature of the body: 59°C
Time taken to cool from 61°C to 59°C: 4 minutes
Explanation:
The rate at which a body cools is directly proportional to the temperature difference between the body and its surroundings. This is given by Newton's Law of Cooling.
Using the formula:
θ(t) = θ_s + (θ_i - θ_s)e^(-kt)
Where:
θ(t) = temperature at time t
θ_s = temperature of surroundings
θ_i = initial temperature of the body
k = cooling constant
t = time
Calculating the cooling constant:
Using the given data, we can find the cooling constant:
59 = 30 + (61 - 30)e^(-4k)
29 = 31e^(-4k)
e^(-4k) = 29/31
-4k = ln(29/31)
k = - ln(29/31) / 4
Finding the time taken to cool from 51°C to 49°C:
Using the formula with the new temperatures:
49 = 30 + (51 - 30)e^(-k*t)
19 = 21e^(-k*t)
e^(-k*t) = 19/21
t = -ln(19/21) / k
Substitute the value of k into the above equation to find t.
After solving, t ≈ 6 minutes.
Therefore, the time taken by the body to cool from 51°C to 49°C will be 6 minutes. Option (b) is correct.
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In a room where the temprature is 30°C, a body cods from 61°C to 59°C in 4 minutes. The time (in min.) taken by the body to cool from 51°C to 49°C will bea)4b)6c)5d)8Correct answer is option 'B'. Can you explain this answer?
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