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Q.
A cup of tea cools from $80^{\circ} C$ to $60^{\circ} C$ in one minute. The ambient temperature is $30^{\circ} C$. In cooling from $60^{\circ} C$ to $50^{\circ} C$. It will take:
Rate of cooling $\alpha$ excess of temperature
$\frac{80^{\circ}-60^{\circ}}{60^{\circ}}=K\left(\frac{80^{\circ}+60^{\circ}}{2}-30^{\circ}\right) $
$\frac{1}{3}=K(4)\,\,\,\,...(1) $
$\frac{60^{\circ}-50^{\circ}}{t}= K \left(\frac{60^{\circ}+50^{\circ}}{2}-30^{\circ}\right)$
$\frac{10}{t}=K(25)\,\,\,\,...(2)$
Dividing equation (1) by (2), we get
$\frac{t}{30}=\frac{40}{25}$
or $t=\frac{40}{25} \times 30$
$=48\, \sec$