A Body Cools in a Room

Q. A body cools from 80 °C to 50 °C in 5 min. The temperature of the surroundings is 20 °C. How long does it take to cool from 60 °C to 30 °C, in minutes? expects 9 min

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Q. A body cools from 80 °C $T_{b} (cf₀) = 80$ °C to 50 °C $T_{b} (cf₁) = 50$ °C in 5 min $Δ t_{cf} = 5$ min. The temperature of the surroundings is 20 °C $T_{r} = 20$ °C. How long does it take to cool from 60 °C $T_{b} (cl₀) = 60$ °C to 30 °C $T_{b} (cl₁) = 30$ °C, in minutes?

The playground

Inputs

first from

T_{b}

80 °C

first to

T_{b}

50 °C

first time

Δ t_{cf}

5 min

room

T_{r}

20 °C

then from

T_{b}

60 °C

then to

T_{b}

30 °C

Outputs

time

Δ t_{cl}

The solution

Solved with NewtonsLawOfCooling

Given

$T_{b} (cf₀) = 80$ °C
$T_{b} (cf₁) = 50$ °C
$Δ t_{cf} = 5$ min
$T_{r} = 20$ °C
$T_{b} (cl₀) = 60$ °C
$T_{b} (cl₁) = 30$ °C

Find $Δ t_{cl}$ .

Working

$T_{b} (cf₀) = 80$ °C $= 353.15$ K
$T_{b} (cf₁) = 50$ °C $= 323.15$ K
$Δ t_{cf} = 5$ min $= 300$ s
$T_{r} = 20$ °C $= 293.15$ K
$T_{b} (cl₀) = 60$ °C $= 333.15$ K
$T_{b} (cl₁) = 30$ °C $= 303.15$ K
cooling goes as the excess over the surroundings(cooling first) $\frac{T_{b} (cf₁) - T_{b} (cf₀)}{Δ t_{cf}} = - k_{c,b} \cdot (\frac{T_{b} (cf₀) + T_{b} (cf₁)}{2} - T_{r})$
$\frac{323.15 - 353.15}{300} = - k_{c,b} \cdot (\frac{353.15 + 323.15}{2} - 293.15)$
$k_{c,b} = 0.00222222222$ s⁻¹
cooling goes as the excess over the surroundings(cooling later) $\frac{T_{b} (cl₁) - T_{b} (cl₀)}{Δ t_{cl}} = - k_{c,b} \cdot (\frac{T_{b} (cl₀) + T_{b} (cl₁)}{2} - T_{r})$
$\frac{303.15 - 333.15}{Δ t_{cl}} = - 0.00222222222 \cdot (\frac{333.15 + 303.15}{2} - 293.15)$
$Δ t_{cl} = 9$ min
$Δ t_{cl} = 540$ s $= 9$ min
$Δ t_{cl} \approx 9$ minto 1 significant figure

Conclusion

The body loses heat at a rate set by how much hotter than its surroundings it is; the same drop nearer room temperature comes more slowly and takes longer.

The Ruby

1require_relative "../physics"23question :a_body_cools_in_a_room do4  kind :numerical5  describes "A body cools from %{first_from} to %{first_to} in %{first_time}. The temperature of the surroundings " \6            "is %{room}. %{time} to cool from %{then_from} to %{then_to}, in minutes?"7  given first_from: "80 °C"8  given first_to: "50 °C"9  given first_time: "5 min"10  given room: "20 °C"11  given then_from: "60 °C"12  given then_to: "30 °C"13  asks time: "How long does it take"14end

1require_relative "../questions/a_body_cools_in_a_room"2require_relative "../science/thermal_properties_of_matter/newtons_law_of_cooling"34solve :a_body_cools_in_a_room, using: NewtonsLawOfCooling do5  objects :body, :room6  body.goes(:cooling_first, in: room)7  body.goes(:cooling_later, in: room)89  take first_from: body.temperature(at: cooling_first.start)10  take first_to: body.temperature(at: cooling_first.end)11  take first_time: cooling_first.time_taken12  take room: room.temperature13  take then_from: body.temperature(at: cooling_later.start)14  take then_to: body.temperature(at: cooling_later.end)1516  time = cooling_later.time_taken17  answer time:, in: "min", figures: 11819  conclusion "The body loses heat at a rate set by how much hotter than its surroundings it is; the same " \20             "drop nearer room temperature comes more slowly and takes longer."21end

1require_relative "../solutions/a_body_cools_in_a_room"23play_with :a_body_cools_in_a_room do4  title "A Body Cools in a Room"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 body during cooling first	$Δ x_{b} (cf)$	m	—
position of body at end of cooling first	$x_{b} (cf₁)$	m	—
position of body at start of cooling first	$x_{b} (cf₀)$	m	—
displacement of body during cooling later	$Δ x_{b} (cl)$	m	—
position of body at end of cooling later	$x_{b} (cl₁)$	m	—
position of body at start of cooling later	$x_{b} (cl₀)$	m	—
average velocity of body during cooling first	$v_{av,b} (cf)$	m/s	—
time taken of cooling first	$Δ t_{cf}$	s	0 … 1e+05
average velocity of body during cooling later	$v_{av,b} (cl)$	m/s	—
time taken of cooling later	$Δ t_{cl}$	s	0 … 1e+05
average speed of body during cooling first	$c_{av,b} (cf)$	m/s	—
path length of body during cooling first	$d_{b} (cf)$	m	—
average speed of body during cooling later	$c_{av,b} (cl)$	m/s	—
path length of body during cooling later	$d_{b} (cl)$	m	—
temperature of body at start of cooling first	$T_{b} (cf₀)$	K	0 … ∞
temperature of body at end of cooling first	$T_{b} (cf₁)$	K	0 … ∞
cooling constant of body	$k_{c,b}$	s⁻¹	—
temperature of room	$T_{r}$	K	0 … ∞
temperature of body at start of cooling later	$T_{b} (cl₀)$	K	0 … ∞
temperature of body at end of cooling later	$T_{b} (cl₁)$	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_{b} (cf) = x_{b} (cf₁) - x_{b} (cf₀)

change of position(cooling first)

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

change of position(cooling later)

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

displacement a time(cooling first)

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

displacement a time(cooling later)

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

path a time(cooling first)

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

path a time(cooling later)

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

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

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

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