Hot Water Cooling From Sixty

solved NewtonsLawOfCooling

Q. Hot water cools from 60 °C $T_{w} (cf₀) = 60$ °C to 50 °C $T_{w} (cf₁) = 50$ °C in the first 10 min $Δ t_{cf} = 10$ min and to 42 °C $T_{w} (cn₁) = 42$ °C in the next $T_{w} (cn₀) = T_{w} (cf₁)$ 10 min $Δ t_{cn} = 10$ min. Then the temperature of the surroundings is, in °C?

Inputs

first from

T_{w}

60 °C

first to

T_{w}

50 °C

first time

Δ t_{cf}

10 min

then to

T_{w}

42 °C

then time

Δ t_{cn}

10 min

Outputs

surroundings

T_{a}

How it's solved

Solved with NewtonsLawOfCooling

Given

$T_{w} (cf₀) = 60$ °C
$T_{w} (cf₁) = 50$ °C
$Δ t_{cf} = 10$ min
$T_{w} (cn₁) = 42$ °C
$Δ t_{cn} = 10$ min
$T_{w} (cn₀) = T_{w} (cf₁)$ from “in the next”

Find $T_{a}$ .

Working

$T_{w} (cf₀) = 60$ °C $= 333.15$ K
$T_{w} (cf₁) = 50$ °C $= 323.15$ K
$Δ t_{cf} = 10$ min $= 600$ s
$T_{w} (cn₁) = 42$ °C $= 315.15$ K
$Δ t_{cn} = 10$ min $= 600$ s
in the next $T_{w} (cn₀) = T_{w} (cf₁)$
$T_{w} (cn₀) = 323.15$
$T_{w} (cn₀) = 323.15$ K
cooling goes as the excess over the surroundings(cooling first) $\frac{T_{w} (cf₁) - T_{w} (cf₀)}{Δ t_{cf}} = - k_{c,w} \cdot (\frac{T_{w} (cf₀) + T_{w} (cf₁)}{2} - T_{a})$
cooling goes as the excess over the surroundings(cooling next) $\frac{T_{w} (cn₁) - T_{w} (cn₀)}{Δ t_{cn}} = - k_{c,w} \cdot (\frac{T_{w} (cn₀) + T_{w} (cn₁)}{2} - T_{a})$
$T_{a} = 10$ °C
$T_{a} = 283.15$ K $= 10$ °C

Conclusion

The water cools faster the hotter it is than its surroundings; the second drop of 8 degrees against the first of 10 puts the surroundings near 10 °C.

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 :hot_water_cooling_from_sixty do4  kind :numerical5  describes "Hot water cools from %{first_from} to %{first_to} in the first %{first_time} and to %{then_to} in the " \6            "next %{then_time}. Then %{surroundings}, in °C?"7  given first_from: "60 °C"8  given first_to: "50 °C"9  given first_time: "10 min"10  given then_to: "42 °C"11  given then_time: "10 min"12  asks surroundings: "the temperature of the surroundings is"13end

solution.rb

1require_relative "../questions/hot_water_cooling_from_sixty"2require_relative "../science/thermal_properties_of_matter/newtons_law_of_cooling"34solve :hot_water_cooling_from_sixty, using: NewtonsLawOfCooling do5  objects :water, :air6  water.goes(:cooling_first, in: air)7  water.goes(:cooling_next, in: air)89  take first_from: water.temperature(at: cooling_first.start)10  take first_to: water.temperature(at: cooling_first.end)11  take first_time: cooling_first.time_taken12  take then_to: water.temperature(at: cooling_next.end)13  take then_time: cooling_next.time_taken14  deduce_from "in the next", water.temperature(at: cooling_next.start) => water.temperature(at: cooling_first.end)1516  surroundings = air.temperature17  answer surroundings:, in: "°C"1819  conclusion "The water cools faster the hotter it is than its surroundings; the second drop of 8 degrees against " \20             "the first of 10 puts the surroundings near 10 °C."21end

playground.rb

1require_relative "../solutions/hot_water_cooling_from_sixty"23play_with :hot_water_cooling_from_sixty do4  title "Hot Water Cooling From Sixty"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 water during cooling first	$Δ x_{w} (cf)$	m	—
position of water at end of cooling first	$x_{w} (cf₁)$	m	—
position of water at start of cooling first	$x_{w} (cf₀)$	m	—
displacement of water during cooling next	$Δ x_{w} (cn)$	m	—
position of water at end of cooling next	$x_{w} (cn₁)$	m	—
position of water at start of cooling next	$x_{w} (cn₀)$	m	—
average velocity of water during cooling first	$v_{av,w} (cf)$	m/s	—
time taken of cooling first	$Δ t_{cf}$	s	0 … 1e+05
average velocity of water during cooling next	$v_{av,w} (cn)$	m/s	—
time taken of cooling next	$Δ t_{cn}$	s	0 … 1e+05
average speed of water during cooling first	$c_{av,w} (cf)$	m/s	—
path length of water during cooling first	$d_{w} (cf)$	m	—
average speed of water during cooling next	$c_{av,w} (cn)$	m/s	—
path length of water during cooling next	$d_{w} (cn)$	m	—
temperature of water at start of cooling first	$T_{w} (cf₀)$	K	0 … ∞
temperature of water at end of cooling first	$T_{w} (cf₁)$	K	0 … ∞
cooling constant of water	$k_{c,w}$	s⁻¹	—
temperature of air	$T_{a}$	K	0 … ∞
temperature of water at start of cooling next	$T_{w} (cn₀)$	K	0 … ∞
temperature of water at end of cooling next	$T_{w} (cn₁)$	K	0 … ∞
cooling constant of air	$k_{c,a}$	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_{w} (cf) = x_{w} (cf₁) - x_{w} (cf₀)

change of position(cooling first)

Δ x_{w} (cn) = x_{w} (cn₁) - x_{w} (cn₀)

change of position(cooling next)

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

displacement a time(cooling first)

v_{av,w} (cn) = \frac{Δ x_{w} (cn)}{Δ t_{cn}}

displacement a time(cooling next)

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

path a time(cooling first)

c_{av,w} (cn) = \frac{d_{w} (cn)}{Δ t_{cn}}

path a time(cooling next)

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

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

\frac{T_{w} (cn₁) - T_{w} (cn₀)}{Δ t_{cn}} = - k_{c,w} \cdot (\frac{T_{w} (cn₀) + T_{w} (cn₁)}{2} - T_{a})

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

T_{w} (cn₀) = T_{w} (cf₁)

in the next