Hot Water Cooling From Sixty

Q. Hot water cools from 60 °C to 50 °C in the first 10 min and to 42 °C in the next 10 min. Then the temperature of the surroundings is, in °C? expects 10 °C

What the draft establishes

Fills in once a draft is checked.

The web of concepts

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The equations it used

None yet.

Check an answer

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?

The playground

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}

The solution

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 Ruby

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

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

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

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