Q1. A body at a temperature of 40ºC is kept in a surrounding of constant temperature of 20ºC. It is observed that its temperature falls to 35ºC in 10 minutes. Find how much more time will it take for the body to attain a temperature of 30ºC.
Question
Q1. A body at a temperature of 40ºC is kept in a surrounding of constant temperature of 20ºC. It is observed that its temperature falls to 35ºC in 10 minutes. Find how much more time will it take for the body to attain a temperature of 30ºC.
Solution
This problem is based on Newton's Law of Cooling, which states that the rate of change of temperature of an object is proportional to the difference between its own temperature and the ambient temperature, i.e., the temperature of its surroundings.
The formula for Newton's Law of Cooling is:
dT/dt = -k(T - Ts)
where:
- dT/dt is the rate of change of temperature
- T is the temperature of the object
- Ts is the temperature of the surroundings
- k is a constant
Given:
- Initial temperature of the body, T1 = 40ºC
- Temperature of the body after 10 minutes, T2 = 35ºC
- Temperature of the surroundings, Ts = 20ºC
- Time taken for the temperature to fall from T1 to T2, t = 10 minutes
We need to find the time it will take for the body to cool from T2 to T3 = 30ºC.
Step 1: Find the value of k We can rearrange the formula to find k:
k = - (1/t) * ln((T2 - Ts) / (T1 - Ts))
Substituting the given values:
k = - (1/10) * ln((35 - 20) / (40 - 20)) k = -0.1 * ln(15 / 20) k = -0.1 * ln(0.75) k ≈ 0.0282
Step 2: Find the time it will take for the body to cool from T2 to T3 We can rearrange the formula to find t:
t = - (1/k) * ln((T3 - Ts) / (T2 - Ts))
Substituting the values:
t = - (1/0.0282) * ln((30 - 20) / (35 - 20)) t = -35.5 * ln(10 / 15) t = -35.5 * ln(0.67) t ≈ 16.5 minutes
Therefore, it will take approximately an additional 16.5 minutes for the body to cool from 35ºC to 30ºC.
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