Question 198: Thermodynamics & KTG

Question

A cup of coffee cools from 90°C to 80°C in t minutes when the room temperature is 20°C. The time taken by the similar cup of coffee to cool from 80°C to 60°C at the same room temperature is:

Accepted Answer

Here, we will use the Newton's law of cooling.
We know, $${{{T_1} - {T_2}} \over {\Delta t}} = k\left( {{{{T_1} + {T_2}} \over 2} - {T_s}} \right)$$ .... (1)
where, T$_1$ and T$_2$ are initial and final temperatures respectively.
$\mathrm{T_s}$ = surrounding temperature
$\Delta t$ = time taken
$k$ = positive constant
Given,
Ist case :
$${T_1} = 90^\circ C$$
$${T_2} = 80^\circ C$$
$$\Delta t = t$$, $${T_s} = 20^\circ C$$
IInd case :
$${T_1} = 80^\circ C$$
$${T_2} = 60^\circ C$$
$${T_s} = 20^\circ C$$, $$\Delta t = ?$$
Now, for Ist case,
$${{90 - 80} \over t} = k\left( {{{90 + 80} \over 2} - 20} \right)$$ ...... (from (1))
$$ \Rightarrow {{10} \over t} = k\left( {85 - 20} \right)$$
$$ \Rightarrow {{10} \over t} = 65k$$ .... (2)
For IInd case,
$${{80 - 60} \over {\Delta t}} = k\left( {{{80 + 60} \over 2} - 20} \right)$$ .... (From (1))
$$ \Rightarrow {{20} \over {\Delta t}} = k\left( {70 - 20} \right)$$
$$ \Rightarrow {{20} \over {\Delta t}} = 50k$$ .... (3)
Now, by (2) $\div$ (3),
$${{{{10} \over t}} \over {{{20} \over {\Delta t}}}} = {{13} \over {10}}$$
$$ \Rightarrow {{\Delta t} \over t}\left( {{1 \over 2}} \right) = {{13} \over 5}$$
$$ \Rightarrow \Delta t = {{13} \over 5}t$$

Answer

$\frac{13}{5}t$

Answer

$\frac{10}{13}t$

Answer

$\frac{5}{13}t$

Answer

$\frac{13}{10}t$

Thermodynamics & KTG

A cup of coffee cools from 90°C to 80°C in t minutes when the room temperature is 20°C. The time taken by the similar cup of coffee to cool from 80°C to 60°C at the same room temperature is:

Choose the correct answer: