Physics · Thermal Properties Of Matter · NEET
Because the rate of cooling depends on how much hotter the body is than the room, not on the size of the drop. When the body is at 80-70 C it is far above room temperature, so it loses heat fast and the 10 C drop happens quickly. When it is at 70-60 C it is closer to the room, so it loses heat slowly and the same 10 C drop takes more time. That is why equal temperature intervals take unequal times.
You use the average of the two temperatures for that stage. Newton's law says the rate depends on the temperature difference from the room, but over an interval we use the average body temperature as a good estimate. For a fall from 80 C to 70 C, the average body temperature is (80 + 70)/2 = 75 C. Then the excess over a 25 C room is 75 - 25 = 50 C. Using the starting value alone is the most common NEET mistake.
You do not need its exact units for these problems. Use the first cooling stage where both the drop and the time are given. Write (T1 - T2)/t = k x (average - room). Plug in the numbers and solve for k. For example 80 to 70 C in 12 minutes with a 25 C room: 10/12 = k x (75 - 25) = 50k, so k = 1/60 per minute. Then reuse this same k for the second stage.
Yes, and it is faster. Divide the equation for stage 1 by the equation for stage 2. The k cancels out. You get (drop1/time1) / (drop2/time2) = (average1 - room)/(average2 - room). Rearrange to find the unknown time. This ratio method avoids fractions and is safest under exam time pressure.
For a single body cooling in two stages, mass, specific heat and area are already baked into the constant k. Since it is the same body cooling in the same room, k stays the same for both stages, so it cancels. You never need to know the mass. Only the temperatures and one known time matter.
An object kept in a large room (air temperature 25 C) takes 12 minutes to cool from 80 C to 70 C. The time taken by the same object to cool from 70 C to 60 C would be nearly:
A cup of coffee cools from 90 C to 80 C in t minutes when the room temperature is 20 C. The time taken by a similar cup of coffee to cool from 80 C to 60 C at the same room temperature (20 C) is:
A body cools from a temperature 3T to 2T in 10 minutes. The surroundings are at temperature T. Assuming Newton's law of cooling holds, the temperature of the body at the end of the next 10 minutes will be:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
(T1 - T2)/t = k x ((T1 + T2)/2 - T0), where T1 and T2 are the start and end temperatures of the stage, t is the time, T0 is the room temperature and k is a constant for that body and room. Find k from a known stage, then reuse it.
The exact solution of Newton's law is an exponential curve. For NEET, if the drop is small compared to the excess over the room, the average temperature of the interval is a very good approximation and makes the maths a simple linear equation. Almost all NEET numericals use this average form.
k has units of per minute or per second, but for two-stage problems you rarely need its value or units. You either compute k as a plain number from stage 1 or cancel it entirely by dividing the two stage equations.
Longer. As the body approaches room temperature, the excess over the room shrinks, so the rate of cooling drops. That is why each successive equal temperature interval takes more and more time, approaching room temperature but theoretically never fully reaching it.
When the temperature drop is large compared to the excess over the room, the average approximation becomes rough. Then you should use the exact exponential form of Newton's law. For standard NEET questions the drops are small, so the average method is accepted.
