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The temperature of a body falls from 50°C to 40°C in 10 minutes. If the temperature of the surroundings is 20°C Then temperature of the body after another 10 minutes will be
- a)36.6°C
- b)33.3°C
- c)35°C
- d)30°C
Correct answer is option 'B'. Can you explain this answer?
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The temperature of a body falls from 50°C to 40°C in 10 minute...
Given Data:
Initial temperature of the body = 50°C
Final temperature of the body = 40°C
Temperature of the surroundings = 20°C
Time interval = 10 minutes
Calculation:
Let the temperature of the body after another 10 minutes be x.
Using Newton's Law of Cooling, we have:
Rate of cooling ∝ (Temperature difference between the body and surroundings)
Let ΔT1 be the initial temperature difference = 50 - 20 = 30°C
Let ΔT2 be the final temperature difference = 40 - 20 = 20°C
According to Newton's Law of Cooling:
ΔT1/ΔT2 = e^(-kt1) / e^(-kt2)
where k is the cooling constant and t1, t2 are the respective time intervals.
Using the given data:
30/20 = e^(-k*10) / e^(-k*20)
Solving the above equation, we get:
k = ln(3/2) / 10
Now, we can find the temperature of the body after another 10 minutes:
x = 20 + (40 - 20) * e^(-ln(3/2)/10 * 10)
x = 20 + 20 * 1/2
x = 20 + 10
x = 30°C
Therefore, the temperature of the body after another 10 minutes will be 30°C, which corresponds to option 'd'.
Initial temperature of the body = 50°C
Final temperature of the body = 40°C
Temperature of the surroundings = 20°C
Time interval = 10 minutes
Calculation:
Let the temperature of the body after another 10 minutes be x.
Using Newton's Law of Cooling, we have:
Rate of cooling ∝ (Temperature difference between the body and surroundings)
Let ΔT1 be the initial temperature difference = 50 - 20 = 30°C
Let ΔT2 be the final temperature difference = 40 - 20 = 20°C
According to Newton's Law of Cooling:
ΔT1/ΔT2 = e^(-kt1) / e^(-kt2)
where k is the cooling constant and t1, t2 are the respective time intervals.
Using the given data:
30/20 = e^(-k*10) / e^(-k*20)
Solving the above equation, we get:
k = ln(3/2) / 10
Now, we can find the temperature of the body after another 10 minutes:
x = 20 + (40 - 20) * e^(-ln(3/2)/10 * 10)
x = 20 + 20 * 1/2
x = 20 + 10
x = 30°C
Therefore, the temperature of the body after another 10 minutes will be 30°C, which corresponds to option 'd'.
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The temperature of a body falls from 50°C to 40°C in 10 minutes. If the temperature of the surroundings is 20°C Then temperature of the body after another 10 minutes will bea)36.6°Cb)33.3°Cc)35°Cd)30°CCorrect answer is option 'B'. Can you explain this answer?
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