Question:

In an experiment to verify Newton’s law of cooling, a graph is plotted between, the temperature difference (ΔT) of the water and surroundings and time as shown in figure. The initial temperature of water is taken as 80ºC. The value of \(t_2\) as mentioned in the graph will be ______.
graph is plotted between, the temperature difference (ΔT) of the water and surroundings and time

Updated On: Mar 19, 2025
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Correct Answer: 16

Solution and Explanation

Temperature of surrounding \(=\) \(20ºC\)
For \(0\) to \(6\) minutes, average temp. \(=\) \(70ºC\)
Rate of cooling \(∝70ºC – 20ºC = 50ºC\)
For \(6\) to \(t_2\) minutes, average temp. \(= 50ºC\)
Rate of cooling \(∝30ºC\)
\(⇒ t_2-6 = \frac 53\) (\(6\) minutes)
\(⇒ t_2= 16 \) minutes

So, the answer is \(16\) minutes.

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Concepts Used:

Newton’s Law of Cooling

Newton’s law of cooling states that the rate of heat loss from a body is directly proportional to the difference in temperature between the body and its surroundings. 

Derivation of Newton’s Law of Cooling

Let a body of mass m, with specific heat capacity s, is at temperature T2 and T1 is the temperature of the surroundings. 

If the temperature falls by a small amount dT2 in time dt, then the amount of heat lost is,

dQ = ms dT2

The rate of loss of heat is given by,

dQ/dt = ms (dT2/dt)                                                                                                                                                                              ……..(2)

Compare the equations (1) and (2) as,

– ms (dT2/dt) = k (T2 – T1)

Rearrange the above equation as:

dT2/(T2–T1) = – (k / ms) dt

dT2 /(T2 – T1) = – Kdt 

where K = k/m s

Integrating the above expression as,

loge (T2 – T1) = – K t + c

or 

T2 = T1 + C’ e–Kt

where C’ = ec