Question

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

Answer 1

Answer:

1 minute

Explanation:

°C cooled from 94°C to 86°C= 94 - 86

= 12°C

Time taken to cool 12°C = 2 minutes

Time taken to cool 1°C = 12÷ 2

= 6 minutes

°C cooled from 71°C to 69°C= 71- 69

= 9° C

Time taken to cool 1°C = 1/9

Time taken to cool 1/9°C = 1/9÷ 9

= 1 minute

Answer 2

Answer:

Time :- 0.7 min = 42 second.

Explanation:

The Average temprature of 94°C and 86°C, above room temprature which is 70°C. Under these Conditions the Pan Cools 8°C in 2 minutes.

Using an Equation, we have

$\frac{Change \: in \: temprature}{Time }$= KΔT

$\frac{8°C}{2 \: min} = K \: (50°C)$

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

$\frac{2°C}{Time} = k(50°C)$

Whwn we duvuse above two equations, we have

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

Time = 0.7 min.

= 42 secinds.