A hot ball cools from 90°C to 10°C in 5 minutes. If the surrounding temperature is 20°C, what is the time taken to cool from 60°C to 30°C?

Asked by Abhisek | 1 year ago |  382

1 Answer

Solution :-

Using Newton’s law of cooling, the cooling rate is directly proportional to the difference in temperature.

Here, average of 90°C and 40°C = 50°C

Surrounding temperature = 20°C

Difference = 50 – 20= 30°C

Under the given conditions, the ball cools 80° C in 5 minutes

Therefore, \( \dfrac{Change\, in\, temperature}{Time} \) 

\( = k\Delta t\)

\( = \dfrac{30}{5} = K × 30\) . . . .  . . . . . . (1)

Where the value of  K is a constant.

The average of 60°C and 30°C = 45°C

⇒  45°C – 20°C = 25°C above the room temperature 

and the body cools by 30°C [ 60°C – 30°C ] within time t ( Assume )

Therefore,\( \dfrac{30}{t}\) = K × 25 . . . . . . . (2)

Dividing equation (1)  by (2), we have:

\( \dfrac{\dfrac{30}{5}}{\dfrac{30}{t}}=\dfrac{K \times 30}{K\times 25}​​= \dfrac{t}{5}= 1.2 \)

Therefore, t = 6 mins.

Answered by Pragya Singh | 1 year ago

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