Question

A cup of tea cools from 80^0C to 60⁰C in one minute. The ambient temperature is 30^0C. In cooling from 60^0C to 50^0C it will take

• A. 30 Seconds
• B. 60 Seconds
• C. 90 Seconds
• D. 50 Seconds

Answer 1

Answer: (D)

50 Seconds

Answer 2

According to Newton's law of cooling

$\frac{\theta_{1} - \theta_{2}}{t} \propto \left\lbrack \frac{\theta_{1} + \theta_{2}}{2} - \theta \right\rbrack$

For first condition $\frac{80 - 60}{60} \propto \left\lbrack \frac{80 + 60}{2} - 30 \right\rbrack$ ..(i)

and for second condition $\frac{60 - 50}{t} \propto \left\lbrack \frac{60 + 50}{2} - 30 \right\rbrack$….(ii)

By solving (i) and (ii) we get t = 48 sec $\widetilde{–}50$sec.