JEE Main 2014 · 12 Apr · Shift 1 · Physics Q11

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PhysicsSingle correct+4 / -1Question 11 of 90

Hot water cools from 60°C to 50°C in the first 10 minutes and to 42°C in the next 10 minutes. The temperature of the surroundings is :

25°C

10°C

15°C

20°C

Answer and worked solution

Official answer: B

Using the average-temperature form of Newton's law with surroundings $\theta$ :
$\frac{10}{10}=k(55-\theta),\qquad\frac8{10}=k(46-\theta).$
Dividing: $\frac{55-\theta}{46-\theta}=\frac54$ , so $\theta=10^\circ$ C.

Topics:Newton's law of cooling

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JEE Main 2014 · 12 Apr · Shift 1, Physics question 11