Question 770: Thermodynamics & KTG

Question

The temperature of a liquid falls from 365 K to 359 K in 3 minutes. The time during which temperature of this liquid falls from 342 K to 338 K is [Let the room temperature be 296 K ]

Accepted Answer

To determine the time it takes for the temperature of a liquid to fall from 342 K to 338 K, we use Newton's law of cooling:
$ \frac{T_1 - T_2}{t} = K\left(\frac{T_1 + T_2}{2} - T_0ight) $
where $ T_0 $ is the room temperature.
Given:
First scenario: Temperature falls from 365 K to 359 K in 3 minutes.
Using the equation:
$ \frac{365 - 359}{3} = K\left[\frac{365 + 359}{2} - 296ight] $
Simplifying:
$ \frac{6}{3} = K\left[\frac{724}{2} - 296ight] $
$ 2 = K(362 - 296) $
$ K = \frac{1}{33} $
Second scenario:
Temperature falls from 342 K to 338 K in $ t $ minutes.
Using the equation:
$ \frac{342 - 338}{t} = \frac{1}{33}\left[\frac{342 + 338}{2} - 296ight] $
Simplifying:
$ \frac{4}{t} = \frac{1}{33}\left[\frac{680}{2} - 296ight] $
$ \frac{4}{t} = \frac{4}{3} $
Solving for $ t $:
$ t = 3 	ext{ minutes} $
Therefore, the time required for the temperature to drop from 342 K to 338 K is 3 minutes.

Answer

6 min

Answer

4 min

Answer

3 min

Answer

2 min

Thermodynamics & KTG

The temperature of a liquid falls from 365 K to 359 K in 3 minutes. The time during which temperature of this liquid falls from 342 K to 338 K is [Let the room temperature be 296 K ]

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