MHT CET · Physics · Thermal Properties of Matter
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 ]
- A 6 min
- B 4 min
- C 3 min
- D 2 min
Answer & Solution
Correct Answer
(C) 3 min
Step-by-step Solution
Detailed explanation
From Newton's law of cooling,
\(\therefore \quad \frac{\mathrm{T}_1-\mathrm{T}_2}{\mathrm{t}}=\mathrm{K}\left(\frac{\mathrm{~T}_1+\mathrm{T}_2}{2}-\mathrm{T}_0\right)\)
where \(T_0\) is the room temperature.
For decrease from 365 K to 359 K ,
\(\begin{aligned}
& \therefore \quad \frac{365-359}{3}=\mathrm{K}\left[\frac{365+359}{2}-296\right] \\
& \quad \mathrm{K}=\frac{1}{33}
\end{aligned}\)
For decrease from 342 K to 338 K ,
\(\frac{342-338}{t}=\frac{1}{33}\left[\frac{342+338}{2}-296\right]\)
\(\begin{aligned} & \frac{4}{t}=\frac{4}{3} \\ \therefore \quad & t=3 \mathrm{~min}\end{aligned}\)
\(\therefore \quad \frac{\mathrm{T}_1-\mathrm{T}_2}{\mathrm{t}}=\mathrm{K}\left(\frac{\mathrm{~T}_1+\mathrm{T}_2}{2}-\mathrm{T}_0\right)\)
where \(T_0\) is the room temperature.
For decrease from 365 K to 359 K ,
\(\begin{aligned}
& \therefore \quad \frac{365-359}{3}=\mathrm{K}\left[\frac{365+359}{2}-296\right] \\
& \quad \mathrm{K}=\frac{1}{33}
\end{aligned}\)
For decrease from 342 K to 338 K ,
\(\frac{342-338}{t}=\frac{1}{33}\left[\frac{342+338}{2}-296\right]\)
\(\begin{aligned} & \frac{4}{t}=\frac{4}{3} \\ \therefore \quad & t=3 \mathrm{~min}\end{aligned}\)
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