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A body takes 5 minutes for cooling from 50°C to 40°C. Its temperature comes down to 33.33°C in next 5 minutes. Temperature of surroundings is
- a)15°C
- b)20°C
- c)25°C
- d)10°C
Correct answer is option 'B'. Can you explain this answer?
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A body takes 5 minutes for cooling from 50°C to 40°C. Its temp...
°C to 40°C in a room where the temperature is 30°C. How much time will it take for the body to cool from 40°C to 30°C?
We can use Newton's law of cooling, which states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings.
Let's denote the temperature of the body at time t as T(t). Then we have:
dT/dt = k(T - 30)
where k is a constant of proportionality.
We know that when T = 50, the body cools from 50°C to 40°C in 5 minutes. This means that:
dT/dt = k(50 - 30) = 20k
and
T(5) - 40 = (50 - 40)e^(-5k)
We can use these two equations to solve for k:
20k = (T(5) - 40)/(e^(-5k))
20k = (50 - 40)e^(5k)
e^(5k) = 1/2
5k = ln(1/2)
k = -ln(2)/5
Now we can use this value of k to find the time it takes for the body to cool from 40°C to 30°C:
dT/dt = -ln(2)/5 (40 - 30) = -ln(2)
Integrating both sides from T = 40 to T = 30 and t = 0 to t = T, we get:
ln(40 - 30) - ln(T - 30) = -ln(2)t
ln(4) - ln(T - 30) = -ln(2)t
ln(T - 30) = ln(4) + ln(2)t
T - 30 = 4e^(2t)
T = 4e^(2t) + 30
We want to find the time when T = 30.
4e^(2t) + 30 = 30
e^(2t) = 0
This equation has no real solutions, which means that the body will never reach a temperature of 30°C. We can see from the equation for T(t) that as t approaches infinity, T(t) approaches 30°C but never reaches it.
Therefore, the answer to the question is that the body will never cool to 30°C in this room.
We can use Newton's law of cooling, which states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings.
Let's denote the temperature of the body at time t as T(t). Then we have:
dT/dt = k(T - 30)
where k is a constant of proportionality.
We know that when T = 50, the body cools from 50°C to 40°C in 5 minutes. This means that:
dT/dt = k(50 - 30) = 20k
and
T(5) - 40 = (50 - 40)e^(-5k)
We can use these two equations to solve for k:
20k = (T(5) - 40)/(e^(-5k))
20k = (50 - 40)e^(5k)
e^(5k) = 1/2
5k = ln(1/2)
k = -ln(2)/5
Now we can use this value of k to find the time it takes for the body to cool from 40°C to 30°C:
dT/dt = -ln(2)/5 (40 - 30) = -ln(2)
Integrating both sides from T = 40 to T = 30 and t = 0 to t = T, we get:
ln(40 - 30) - ln(T - 30) = -ln(2)t
ln(4) - ln(T - 30) = -ln(2)t
ln(T - 30) = ln(4) + ln(2)t
T - 30 = 4e^(2t)
T = 4e^(2t) + 30
We want to find the time when T = 30.
4e^(2t) + 30 = 30
e^(2t) = 0
This equation has no real solutions, which means that the body will never reach a temperature of 30°C. We can see from the equation for T(t) that as t approaches infinity, T(t) approaches 30°C but never reaches it.
Therefore, the answer to the question is that the body will never cool to 30°C in this room.
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A body takes 5 minutes for cooling from 50°C to 40°C. Its temperature comes down to 33.33°C in next 5 minutes. Temperature of surroundings isa)15°Cb)20°Cc)25°Cd)10°CCorrect answer is option 'B'. Can you explain this answer?
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