Thermal Properties of Matter

JEE Main level, single correct, medium

118 of 126

Question 118 of 126

A body cools in 77 minutes from 60oC{60}^{o}C to 40oC{40}^{o}C. The temperature of the surrounding is 10oC{10}^{o}C. The temperature of the body after the next 77 minutes will be

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Answer: (D) 28oC{28}^{o}C

Newton's law of cooling states that
(T1−T2)t=K((T1+T22)−T0)...(i)\dfrac{\left({T}_{1}-{T}_{2}\right)}{t}=K\left(\left(\dfrac{{T}_{1}+{T}_{2}}{2}\right)-{T}_{0}\right)...(i)
where T0{T}_{0} is the temperature of the surrounding. From equation (i),
60−407×60=K(60+402−10)\dfrac{60-40}{7\times 60}=K\left(\dfrac{60+40}{2}-10\right)
⇒207×60=40K⇒K=207×60×40\Rightarrow \dfrac{20}{7\times 60}=40K \Rightarrow K=\dfrac{20}{7\times 60\times 40}
Let the temperature of the body after the given time be x∘Cx^{\circ}C. By Newton's law of cooling,
40−x7×60=K(40+x2−10)⇒40−x7×60=114×60(20+x2)⇒160−4x=20+x⇒x=28∘C\dfrac{40-x}{7\times 60}=K\left(\dfrac{40+x}{2}-10\right) \Rightarrow \dfrac{40-x}{7\times 60}=\dfrac{1}{14\times 60}(\dfrac{20+x}{2}) \Rightarrow 160-4x=20+x \Rightarrow x=28^{\circ}C

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