Question

A liquid takes 5 minutes to cool from ${80}^{0}Cto{50}^{0}C$ . How much time will it take to cool from ${60}^{0}Cto{30}^{0}C$ ? The temperature of the surrounding is ${20}^0$ C .

• A. 5 minutes
• B. 9 minutes
• C. 4 minutes
• D. 12 minutes

Answer 1

The correct option is B

9 minutes

According to Newton’s law of cooling, $\left(\frac{{\theta }_1-{\theta }_2}{t}\right)=K\left[(\frac{{\theta }_1+{\theta }_2}{2}-{\theta }_0)\right]Sothat\left(\frac{80-50}{5}\right)=K\left[(\frac{80+50}{2}-20)\right]And\left(\frac{60-30}{t}\right)=K[\left(\frac{60+30}{2}\right)-20]$

Solving these for t, we get t=9 minute.