Question

A pan filled with hot food cools from 94°C to 86°C in 2 minutes when the room temperature is at 20°C. How long will it take to cool from 71° C to 69° C?​

Answer 1

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Explanation:

➣ The average temperature of 94° C and 86°C is 90° C, which is 70° C above the room temperature. Under these conditions the pan cools 8°C in 2 Minutes.

$\frac{Change \: in \: temperature \: }{Time} = K \triangle T$

$\frac{8 \: °C}{2min} = K(70 \: °C)$

➣The average of 69° C and 71° C is 70° C, which is 50° C above room temperature. K is the same for this situation as for the original.

$\frac{2 \: °C}{Time \: } = K(50 \: °C)$

When we divide above two equations, we have

$\frac{8 \: °C /2min}{2 \: °C / \: time \: } = \frac{K(70 \: °C)}{K(50 \: °C)}$

Time = 0.7 min

= 42s

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