Question

The rate at which a substance cools in moving air is proportional to the difference between the temperatures of the substance and that of the air. If the temperature of the air is 290⁰ K and the substance cools from 370⁰ K to 330⁰K in 10 minutes, when will the temperature be 295⁰K –

• A. 45 minutes
• B. 30 minutes
• C. 35 minutes
• D. 40 minutes

Answer 1

Answer: (D)

40 minutes

Answer 2

Let T be the temperature of the substance at a time t then – $\frac{dT}{dt}$ µ (T – 290) Ž $\frac{dT}{dt}$ = – k (T – 290)

Where k is constant of proportionality and negative sign denote rate of cooling.

or $\frac{dT}{(T - 290)}$ = – k dt

integrating, we get

$\int_{}^{}\frac{dT}{(T - 290)}$ = – k $\int_{}^{}{dt}$ Ž ln (T – 290) = – kt + ln c

Ž $\left( \frac{T - 290}{c} \right)$ = e^–kt

or (T – 290) = ce^–kt

If initially i.e., when t = 0 & T = 370

Then (370 – 290) = ce

\ c = 80

\ T – 290 = 80e ^k... (1)

and for t = 10, T = 330

\ 330 – 290 = 80e^–10k Ž (40) = 80e^–10k

Ž 2 = e^10k

ln2 = 10 k ... (2)

To find t, when T = 295

From (1), 295 – 290 = 80e^–kt

Ž$\frac{5}{80}$= e^–kt Ž ln 16 = kt

or 4 ln 2 = kt ... (3)

Dividing (3) by (2) then 4 = $\frac{t}{10}$

\ t = 40 minutes