Question

A plate is removed from an oven at temperature of 50◦ and placed in a large room at 20◦ . After 5 minutes the temperature of the plate is 40◦ . How much longer will it take to reach 30◦ ?

Answer 1

Step-by-step explanation:

The given data is as follows.

Initial temperature (
`T_(1)`) =
`50^(o)C`

Room temperature (
`T_(a)`) =
`20^(o)C`

Time (t) = 5 min

Final temperature (T) =
`40^(o)C`

According to the unsteady state equation,

`(ln (T - T_(a)))/((T_(1) - T_(a))) = (-h * A * t)/(dV * C)`

where, h = heat transfer coefficient

A = area

d = density

V = volume

C = specific heat capacity

All of these are constants and can be expressed as K

Therefore, K =
`(h * A)/(dV * C)`

`(ln (T - T_(a)))/((T_(1) - T_(a))) = -K * t`

`(ln (40 - 20))/((50 - 20)) = -K * 5` 0.4055 = 5K

K = 0.0811

After 5 minutes, the new temperature will be as follows.

New temperature (
`T_(2)`) =
`30^(o)C`

Time (t) = ?

Again, from the unsteady state conduction,

`(ln (T_(2) - T_(a)))/((T_(i) - T_(a))) = [tex](-h * A * t)/(dV * C)` =
`-K * t`

`(ln [(30 - 20))/((50 - 20)) = - 0.0811 * t`

1.0986 =
`0.0811 * t`

t = 13.55 min

Thus, we can conclude that after 13.55 minutes the temperature reaches to
`30^(o)C`.