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A bowl filled with very hot soup cools from $98^{\circ}\,C$ to $86^{\circ}\,C$ in $2$ minutes when the room temperature is $22^{\circ}\,C$. $..........\,minutes$ long it will take to cool from $75^{\circ}\,C$ to $69^{\circ}\,C$ ?
$2$
$1.4$
$0.5$
$1$
Solution
$\frac{\Delta Q }{\Delta t }=- K \left( T – T _0\right)$
$\frac{\Delta Q }{\Delta t }=- K \left( T _{ avg }- T _0\right)$
$(i)$ $\frac{ ms \times 12}{2}=- K \left(\frac{98+86}{2}-22\right)$
$6=-\frac{ K }{ ms }\left[\frac{98+86}{2}-22\right]$
$6=-\frac{ K }{ ms }[70]……….(i)$
$(ii)$ $\frac{ ms \times 6}{\Delta t }=- K \left(\frac{75+69}{2}-22\right)$ $\frac{6}{\Delta t }=-\frac{ K }{ ms }(50) \ldots \ldots \ldots \ldots$ (ii)
$(ii)$ $\div$ $(i)$
$\frac{6}{\Delta t (6)}=\frac{50}{70}$
$\Delta t =\frac{7}{5}=1.4\,min$
