A Pan of Hot Food

Q. A pan filled with hot food cools from 94 °C to 86 °C in 2 min when the room temperature is at 20 °C. How long will it take to cool from 71 °C to 69 °C, in seconds? expects 42 s

What the draft establishes

Fills in once a draft is checked.

The web of concepts

given read from the words equation asked

The equations it used

None yet.

Check an answer

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?

The playground

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}

The solution

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 Ruby

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

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

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

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)