Question

4MUM
52. A body cools from 67°C to 37°C. If this takes time t when the surrounding temperature is 27°C,
what will be the time taken if the surrounding temperature is 7°C?
A 21
B. t/3
C. 1/2
D. t/4
ths.
made of the same material and having the same area of cross-section
please give me a detailed solution​

Answer 1

Answer:

$\boxed{\mathfrak{(c) \ \frac{t}{2}}}$

Explanation:

Initial temperature ($\rm T_i$) = 67°C

Final temperature ($\rm T_f$) = 37°C

Time taken for cooling when surrounding temperature is 27°C = t

Let time taken for cooling when surrounding temperature is 7°C be t'

From Netwon's Law of Cooling we have:

$\boxed{ \bold{\frac{T_i - T_f}{t} \propto (\frac{T_i + T_f}{2} - T_o)}}$

$\rm T_o \longrightarrow$ Surrounding temperature

So, by using Netwon's Law of Cooling we get:

$\rm \implies \dfrac{67 - 37 }{t} \propto (\dfrac{67 + 37 }{2} - 27 ) \: \: \: \: \: ...eq_1 \\ \\ \rm \implies \dfrac{67 - 37 }{t'} \propto (\dfrac{67 + 37 }{2} - 7 ) \: \: \: \: \: ...eq_2$

Dividing $\rm eq_1$ by $\rm eq_2$ we get:

$\rm \implies \dfrac{\dfrac{ \cancel{67 - 37 }}{t} }{\dfrac{ \cancel{67 - 37} }{t'} } = \dfrac{(\dfrac{67 + 37 }{2} - 27 )}{(\dfrac{67 + 37 }{2} - 7 )} \\ \\ \rm \implies \dfrac{t'}{t} = \dfrac{52 - 27}{52 - 7} \\ \\ \rm \implies \dfrac{t'}{t} = d\frac{25}{45} \\ \\ \rm \implies \dfrac{t'}{t} \approx \dfrac{1}{2} \\ \\ \rm \implies t' \approx \frac{t}{2}$