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)

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