A Pan of Hot Food

solved NewtonsLawOfCooling

Q. A pan filled with hot food cools from 94 °C $T_{p} (cf₀) = 94$ °C to 86 °C $T_{p} (cf₁) = 86$ °C in 2 min $Δ t_{cf} = 2$ min when the room temperature is at 20 °C $T_{r} = 20$ °C. How long will it take to cool from 71 °C $T_{p} (cl₀) = 71$ °C to 69 °C $T_{p} (cl₁) = 69$ °C, in seconds?

Inputs

first from

T_{p}

94 °C

first to

T_{p}

86 °C

first time

Δ t_{cf}

2 min

room

T_{r}

20 °C

then from

T_{p}

71 °C

then to

T_{p}

69 °C

Outputs

time

Δ t_{cl}

How it's solved

Solved with NewtonsLawOfCooling

Given

$T_{p} (cf₀) = 94$ °C
$T_{p} (cf₁) = 86$ °C
$Δ t_{cf} = 2$ min
$T_{r} = 20$ °C
$T_{p} (cl₀) = 71$ °C
$T_{p} (cl₁) = 69$ °C

Find $Δ t_{cl}$ .

Working

$T_{p} (cf₀) = 94$ °C $= 367.15$ K
$T_{p} (cf₁) = 86$ °C $= 359.15$ K
$Δ t_{cf} = 2$ min $= 120$ s
$T_{r} = 20$ °C $= 293.15$ K
$T_{p} (cl₀) = 71$ °C $= 344.15$ K
$T_{p} (cl₁) = 69$ °C $= 342.15$ K
cooling goes as the excess over the surroundings(cooling first) $\frac{T_{p} (cf₁) - T_{p} (cf₀)}{Δ t_{cf}} = - k_{c,p} \cdot (\frac{T_{p} (cf₀) + T_{p} (cf₁)}{2} - T_{r})$
$\frac{359.15 - 367.15}{120} = - k_{c,p} \cdot (\frac{367.15 + 359.15}{2} - 293.15)$
$k_{c,p} = 9.52380952 \times 10^{−4}$ s⁻¹
cooling goes as the excess over the surroundings(cooling later) $\frac{T_{p} (cl₁) - T_{p} (cl₀)}{Δ t_{cl}} = - k_{c,p} \cdot (\frac{T_{p} (cl₀) + T_{p} (cl₁)}{2} - T_{r})$
$\frac{342.15 - 344.15}{Δ t_{cl}} = - 9.52380952 \times 10^{−4} \cdot (\frac{344.15 + 342.15}{2} - 293.15)$
$Δ t_{cl} = 42.00 s$
$Δ t_{cl} \approx 42$ sto 2 significant figures

Conclusion

The pan loses heat at a rate set by how much hotter than the room it is; nearer the room's temperature it cools more slowly, so a smaller drop still takes a while.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb

1require_relative "../physics"23question :a_pan_of_hot_food do4  kind :numerical5  describes "A pan filled with hot food cools from %{first_from} to %{first_to} in %{first_time} when the room " \6            "temperature is at %{room}. %{time} to cool from %{then_from} to %{then_to}, in seconds?"7  given first_from: "94 °C"8  given first_to: "86 °C"9  given first_time: "2 min"10  given room: "20 °C"11  given then_from: "71 °C"12  given then_to: "69 °C"13  asks time: "How long will it take"14end

solution.rb

1require_relative "../questions/a_pan_of_hot_food"2require_relative "../science/thermal_properties_of_matter/newtons_law_of_cooling"34solve :a_pan_of_hot_food, using: NewtonsLawOfCooling do5  objects :pan, :room6  pan.goes(:cooling_first, in: room)7  pan.goes(:cooling_later, in: room)89  take first_from: pan.temperature(at: cooling_first.start)10  take first_to: pan.temperature(at: cooling_first.end)11  take first_time: cooling_first.time_taken12  take room: room.temperature13  take then_from: pan.temperature(at: cooling_later.start)14  take then_to: pan.temperature(at: cooling_later.end)1516  time = cooling_later.time_taken17  answer time:, in: "s", figures: 21819  conclusion "The pan loses heat at a rate set by how much hotter than the room it is; nearer the room's " \20             "temperature it cools more slowly, so a smaller drop still takes a while."21end

playground.rb

1require_relative "../solutions/a_pan_of_hot_food"23play_with :a_pan_of_hot_food do4  title "A Pan of Hot Food"5end

The laws it drew on

quantity	written	units	within	from
position	$x$	m	—	Velocity
displacement	$Δ x$	m	—	Velocity
path length	$d$	m	—	Velocity
time taken	$Δ t$	s	0 … 1e+05	Velocity
average velocity	$v_{av}$	m/s	—	Velocity
average speed	$c_{av}$	m/s	—	Velocity
starting position	$x_{0}$	m	—	Velocity
final position	$x$	m	—	Velocity
temperature	$T$	K	0 … ∞	NewtonsLawOfCooling
cooling constant	$k_{c}$	s⁻¹	—	NewtonsLawOfCooling
starting temperature	$T_{0}$	K	0 … ∞	NewtonsLawOfCooling
final temperature	$T_{f}$	K	0 … ∞	NewtonsLawOfCooling
displacement of pan during cooling first	$Δ x_{p} (cf)$	m	—
position of pan at end of cooling first	$x_{p} (cf₁)$	m	—
position of pan at start of cooling first	$x_{p} (cf₀)$	m	—
displacement of pan during cooling later	$Δ x_{p} (cl)$	m	—
position of pan at end of cooling later	$x_{p} (cl₁)$	m	—
position of pan at start of cooling later	$x_{p} (cl₀)$	m	—
average velocity of pan during cooling first	$v_{av,p} (cf)$	m/s	—
time taken of cooling first	$Δ t_{cf}$	s	0 … 1e+05
average velocity of pan during cooling later	$v_{av,p} (cl)$	m/s	—
time taken of cooling later	$Δ t_{cl}$	s	0 … 1e+05
average speed of pan during cooling first	$c_{av,p} (cf)$	m/s	—
path length of pan during cooling first	$d_{p} (cf)$	m	—
average speed of pan during cooling later	$c_{av,p} (cl)$	m/s	—
path length of pan during cooling later	$d_{p} (cl)$	m	—
temperature of pan at start of cooling first	$T_{p} (cf₀)$	K	0 … ∞
temperature of pan at end of cooling first	$T_{p} (cf₁)$	K	0 … ∞
cooling constant of pan	$k_{c,p}$	s⁻¹	—
temperature of room	$T_{r}$	K	0 … ∞
temperature of pan at start of cooling later	$T_{p} (cl₀)$	K	0 … ∞
temperature of pan at end of cooling later	$T_{p} (cl₁)$	K	0 … ∞
cooling constant of room	$k_{c,r}$	s⁻¹	—

x_{0} \equiv x_{i} (start of motion)

starting position

x \equiv x_{i} (end of motion)

final position

T_{0} \equiv T_{i} (start of motion)

starting temperature

T_{f} \equiv T_{i} (end of motion)

final temperature

d_{i} (whole) = d_{i} (part 1) + d_{i} (part 2)

path in parts

Δ t_{w} = Δ t_{p} + Δ t_{p}

time in parts

\frac{T_{i} (end of cooling) - T_{i} (start of cooling)}{Δ t_{c}} = - k_{c,i} \cdot (\frac{T_{i} (start of cooling) + T_{i} (end of cooling)}{2} - T_{s})

cooling goes as the excess over the surroundings

Δ x_{p} (cf) = x_{p} (cf₁) - x_{p} (cf₀)

change of position(cooling first)

Δ x_{p} (cl) = x_{p} (cl₁) - x_{p} (cl₀)

change of position(cooling later)

v_{av,p} (cf) = \frac{Δ x_{p} (cf)}{Δ t_{cf}}

displacement a time(cooling first)

v_{av,p} (cl) = \frac{Δ x_{p} (cl)}{Δ t_{cl}}

displacement a time(cooling later)

c_{av,p} (cf) = \frac{d_{p} (cf)}{Δ t_{cf}}

path a time(cooling first)

c_{av,p} (cl) = \frac{d_{p} (cl)}{Δ t_{cl}}

path a time(cooling later)

\frac{T_{p} (cf₁) - T_{p} (cf₀)}{Δ t_{cf}} = - k_{c,p} \cdot (\frac{T_{p} (cf₀) + T_{p} (cf₁)}{2} - T_{r})

cooling goes as the excess over the surroundings(cooling first)

\frac{T_{p} (cl₁) - T_{p} (cl₀)}{Δ t_{cl}} = - k_{c,p} \cdot (\frac{T_{p} (cl₀) + T_{p} (cl₁)}{2} - T_{r})

cooling goes as the excess over the surroundings(cooling later)