In 5 minutes, a body cools from $75^{\circ} \mathrm{C}$ to $65^{\circ} \mathrm{C}$ at room temperature of $25^{\circ} \mathrm{C}$. The temperature of body at the end of next 5 minutes is ______ ${ }^{\circ} \mathrm{C}$.

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Heat & Thermodynamics

In 5 minutes, a body cools from

7 5^{\circ} C

6 5^{\circ} C

at room temperature of

2 5^{\circ} C

. The temperature of body at the end of next 5 minutes is ______

^{\circ} C

Answer: 57

Solution

Newton's law of cooling, applied over a short interval with the mean temperature of the interval as the body temperature, gives

\frac{θ _{1} - θ _{2}}{t} = k (\frac{θ _{1} + θ _{2}}{2} - θ_{0})

For the first five minutes, with

θ_{0} = 25^{\circ} C

\frac{75 - 65}{5} = k (\frac{75 + 65}{2} - 25) = 45 k \Rightarrow k = \frac{2}{45} mi n^{- 1}

For the next five minutes, ending at

θ

\frac{65 - θ}{5} = \frac{2}{45} (\frac{65 + θ}{2} - 25)

9 (65 - θ) = 15 + θ \Rightarrow 10 θ = 570

The body is at

57^{\circ} C

at the end of the next five minutes.

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Integer answer

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