Question

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:

• A. 50 sec
• B. 90 sec
• C. 60 sec
• D. 48 sec

Answer 1

Answer: (D)

48 sec

Answer 2

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 $t=\frac{40}{25} \times 30$
$=48\, \sec$