JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Heat & Thermodynamics

In an experment ot verify Newton's law of cooling, a graph is plotted between, the temperature difference

(Δ T)

of the water and surroundings and time as shown in figure. The initial temperature of water is taken as

8 0^{\circ} C

. The value of

t_{2}

as mentioned in the graph will be

____

Answer: 16

Solution

Newton's law of cooling:

T - T_{0} = (T_{i} - T_{0}) e^{- k t}

, where

T_{0}

is the surrounding temperature. From the graph, at

t = 0

Δ T = 60

(so

T_{0} = 2 0^{\circ}

C). At

t = 6

min:

Δ T = 40

, giving

40 = 60 e^{- 6 k}

, so

e^{- 6 k} = 2/3

and

6 k = ln (3/2) = ln 1.5

. At

t = t_{2}

Δ T = 20

, giving

20 = 60 e^{- k t_{2}}

, so

e^{- k t_{2}} = 1/3

and

k t_{2} = ln 3

. Therefore,

\frac{t _{2}}{6} = \frac{l n 3}{l n 1.5} = \frac{l n 3}{l n 3 - l n 2} \approx 2.71

, giving

t_{2} \approx 16.25 \approx 16

min.

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Integer answer