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.

Options

• 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,

`"dT"/"dt" = "k"("T" - "T"_0)`

`"dT"/("T" - "T"_0)` = kdt

`=> int "dT"/("T" - "T"_0) = "k" int "dt"`

`=> log("T" - "T"_0) = "kt" + "C"`

Here, T₀ = 290

∴ log(T - 290) = kt + C

When, t = 0, T = 370

C = log 80

`log(("T" - 290)/80)` = kt

When, t = 10, T = 330

∴ k = `1/10 log (1/2)`

∴ log`(("T" - 290)/80) = "t"/10 log  1/2`

When, T = 295

∴ `log  1/16 = "t"/10 log  1/2`

`=> 10 xx 4 log  1/2 = "t" log  1/2`

⇒ t = 40