Question

The rate at which the metal cools in moving air is proportional to the difference of temperatures between the metal and air. If the air temperature is 290 K and the metal temperature drops from 370 K to 330 K in 1 O min, then the time required to drop the temperature upto 295 K.

विकल्प

• 30 min

• 35 min

• 20 min

• 40 min

Answer 1

40 min

Explanation:

Let T be temperature of metal at any time t and T₀ is temperature of air

According to given condition,

${\displaystyle \frac{\text{dT}}{\text{dt}}=\text{k}(\text{T}-{\text{T}}_0)}$

${\displaystyle \frac{\text{dT}}{\text{T}-{\text{T}}_0}}$ = kdt

${\displaystyle \Rightarrow \int \frac{\text{dT}}{\text{T}-{\text{T}}_0}=\text{k}\int \text{dt}}$

${\displaystyle \Rightarrow \log (\text{T}-{\text{T}}_0)=\text{kt}+\text{C}}$

Here, T₀ = 290

∴ log(T - 290) = kt + C

When, t = 0, T = 370

C = log 80

${\displaystyle \log \left(\frac{\text{T}-290}{80}\right)}$ = kt

When, t = 10, T = 330

∴ k = ${\displaystyle \frac{1}{10}\log \left(\frac{1}{2}\right)}$

∴ log${\displaystyle \left(\frac{\text{T}-290}{80}\right)=\frac{\text{t}}{10}\log \frac{1}{2}}$

When, T = 295

∴ ${\displaystyle \log \frac{1}{16}=\frac{\text{t}}{10}\log \frac{1}{2}}$

${\displaystyle \Rightarrow 10\times 4\log \frac{1}{2}=\text{t}\log \frac{1}{2}}$

⇒ t = 40