Steam Passed Into Water Till Ninety

solved Calorimetry

Q. Steam $T_{1,s} = 100$ °C, $s_{s} = s_{w}$ is passed into 54 g $m_{w} = 54$ g of water $s_{w} = 1$ cal/g °C at 30 °C $T_{1,w} = 30$ °C till the temperature of the mixture $T_{2,s} = T_{2,w}$ becomes 90 °C $T_{2,w} = 90$ °C. If the latent heat of steam is 536 cal/g $L_{s} = 536$ cal/g, the mass of the mixture will be, in grams?

Inputs

water mass

m_{w}

54 g

water temperature

T_{1,w}

30 °C

mixed

T_{2,w}

90 °C

latent heat

L_{s}

536 cal/g

Outputs

total

m_{m}

How it's solved

Solved with Calorimetry

Given

$m_{w} = 54$ g
$T_{1,w} = 30$ °C
$T_{2,w} = 90$ °C
$L_{s} = 536$ cal/g
$s_{w} = 1$ cal/g °C from “of water”
$T_{1,s} = 100$ °C from “Steam”
$s_{s} = s_{w}$ from “Steam”
$T_{2,s} = T_{2,w}$ from “the temperature of the mixture”
$Q_{w} = Q_{L,s} - Q_{s}$ from “Steam is passed into”
$m_{m} = m_{w} + m_{s}$ from “Steam is passed into”

Find $m_{m}$ .

Working

$m_{w} = 54$ g $= 0.054$ kg
$T_{1,w} = 30$ °C $= 303.15$ K
$T_{2,w} = 90$ °C $= 363.15$ K
$L_{s} = 536$ cal/g $= 2.2441248 \times 10^{6}$ J/kg
$s_{w} = 1$ cal/g °C $= 4186.8$ J kg⁻¹ K⁻¹
$T_{1,s} = 100$ °C $= 373.15$ K
Steam $s_{s} = s_{w}$
$s_{s} = 4186.8$
$s_{s} = 4186.8$ J kg⁻¹ K⁻¹
the temperature of the mixture $T_{2,s} = T_{2,w}$
$T_{2,s} = 363.15$
$T_{2,s} = 363.15$ K
heat warms a body(water) $Q_{w} = m_{w} \cdot s_{w} \cdot (T_{2,w} - T_{1,w})$
$Q_{w} = 0.054 \cdot 4186.8 \cdot (363.15 - 303.15)$
$Q_{w} = 13565.232$ J
heat warms a body(steam) $Q_{s} = m_{s} \cdot s_{s} \cdot (T_{2,s} - T_{1,s})$
Steam is passed into (heat to warm of water) $Q_{w} = Q_{L,s} - Q_{s}$
heat changes a phase(steam) $Q_{L,s} = m_{s} \cdot L_{s}$
$m_{s} = 0.00593406593$ kg
Steam is passed into (mass of mixture) $m_{m} = m_{w} + m_{s}$
$m_{m} = 0.054 + 0.00593406593$
$m_{m} = 59.93$ g
$m_{m} = 0.0599340659$ kg $= 59.9340659$ g
$m_{m} \approx 60$ gto 2 significant figures

Conclusion

Each gram of steam gives up its latent heat as it condenses and ten calories more as it cools to 90 °C; the water takes sixty calories a gram to warm, so about six grams of steam join the 54 g.

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 :steam_passed_into_water_till_ninety do4  kind :numerical5  describes "Steam is passed into %{water_mass} of water at %{water_temperature} till the temperature of the " \6            "mixture becomes %{mixed}. If the latent heat of steam is %{latent_heat}, %{total}, in grams?"7  given water_mass: "54 g"8  given water_temperature: "30 °C"9  given mixed: "90 °C"10  given latent_heat: "536 cal/g"11  asks total: "the mass of the mixture will be"12end

solution.rb

1require_relative "../questions/steam_passed_into_water_till_ninety"2require_relative "../science/thermal_properties_of_matter/calorimetry"34solve :steam_passed_into_water_till_ninety, using: Calorimetry do5  objects :steam, :water, :mixture67  take water_mass: water.mass, water_temperature: water.temperature_at_first, mixed: water.temperature_after8  take latent_heat: steam.latent_heat910  deduce_from "of water", water.specific_heat_capacity => "1 cal/g °C"11  deduce_from "Steam", steam.temperature_at_first => "100 °C", steam.specific_heat_capacity => water.specific_heat_capacity12  deduce_from "the temperature of the mixture", steam.temperature_after => water.temperature_after13  deduce_from "Steam is passed into", water.heat_to_warm => steam.heat_to_change_phase - steam.heat_to_warm,14                                      mixture.mass => water.mass + steam.mass1516  total = mixture.mass17  answer total:, in: "g", figures: 21819  conclusion "Each gram of steam gives up its latent heat as it condenses and ten calories more as it cools to " \20             "90 °C; the water takes sixty calories a gram to warm, so about six grams of steam join the 54 g."21end

playground.rb

1require_relative "../solutions/steam_passed_into_water_till_ninety"23play_with :steam_passed_into_water_till_ninety do4  title "Steam Passed Into Water Till Ninety"5end

The laws it drew on

quantity	written	units	within	from
mass	$m$	kg	0 … ∞	Calorimetry
specific heat capacity	$s$	J kg⁻¹ K⁻¹	—	Calorimetry
latent heat	$L$	J/kg	—	Calorimetry
temperature at first	$T_{1}$	K	—	Calorimetry
temperature after	$T_{2}$	K	—	Calorimetry
heat to warm	$Q$	J	—	Calorimetry
heat to change phase	$Q_{L}$	J	—	Calorimetry
heat taken in	$Q_{in}$	J	—	Calorimetry
heat to warm of ground	$Q_{g}$	J	—
mass of ground	$m_{g}$	kg	0 … ∞
specific heat capacity of ground	$s_{g}$	J kg⁻¹ K⁻¹	—
temperature after of ground	$T_{2,g}$	K	—
temperature at first of ground	$T_{1,g}$	K	—
heat to warm of steam	$Q_{s}$	J	—
mass of steam	$m_{s}$	kg	0 … ∞
specific heat capacity of steam	$s_{s}$	J kg⁻¹ K⁻¹	—
temperature after of steam	$T_{2,s}$	K	—
temperature at first of steam	$T_{1,s}$	K	—
heat to warm of water	$Q_{w}$	J	—
mass of water	$m_{w}$	kg	0 … ∞
specific heat capacity of water	$s_{w}$	J kg⁻¹ K⁻¹	—
temperature after of water	$T_{2,w}$	K	—
temperature at first of water	$T_{1,w}$	K	—
heat to warm of mixture	$Q_{m}$	J	—
mass of mixture	$m_{m}$	kg	0 … ∞
specific heat capacity of mixture	$s_{m}$	J kg⁻¹ K⁻¹	—
temperature after of mixture	$T_{2,m}$	K	—
temperature at first of mixture	$T_{1,m}$	K	—
heat to change phase of ground	$Q_{L,g}$	J	—
latent heat of ground	$L_{g}$	J/kg	—
heat to change phase of steam	$Q_{L,s}$	J	—
latent heat of steam	$L_{s}$	J/kg	—
heat to change phase of water	$Q_{L,w}$	J	—
latent heat of water	$L_{w}$	J/kg	—
heat to change phase of mixture	$Q_{L,m}$	J	—
latent heat of mixture	$L_{m}$	J/kg	—
heat taken in of ground	$Q_{in,g}$	J	—
heat taken in of steam	$Q_{in,s}$	J	—
heat taken in of water	$Q_{in,w}$	J	—
heat taken in of mixture	$Q_{in,m}$	J	—

Q_{b} = m_{b} \cdot s_{b} \cdot (T_{2,b} - T_{1,b})

heat warms a body

Q_{L,b} = m_{b} \cdot L_{b}

heat changes a phase

Q_{in,b} = Q_{b} + Q_{L,b}

heat taken in warms and changes phase

Q_{g} = m_{g} \cdot s_{g} \cdot (T_{2,g} - T_{1,g})

heat warms a body(ground)

Q_{s} = m_{s} \cdot s_{s} \cdot (T_{2,s} - T_{1,s})

heat warms a body(steam)

Q_{w} = m_{w} \cdot s_{w} \cdot (T_{2,w} - T_{1,w})

heat warms a body(water)

Q_{m} = m_{m} \cdot s_{m} \cdot (T_{2,m} - T_{1,m})

heat warms a body(mixture)

Q_{L,g} = m_{g} \cdot L_{g}

heat changes a phase(ground)

Q_{L,s} = m_{s} \cdot L_{s}

heat changes a phase(steam)

Q_{L,w} = m_{w} \cdot L_{w}

heat changes a phase(water)

Q_{L,m} = m_{m} \cdot L_{m}

heat changes a phase(mixture)

Q_{in,g} = Q_{g} + Q_{L,g}

heat taken in warms and changes phase(ground)

Q_{in,s} = Q_{s} + Q_{L,s}

heat taken in warms and changes phase(steam)

Q_{in,w} = Q_{w} + Q_{L,w}

heat taken in warms and changes phase(water)

Q_{in,m} = Q_{m} + Q_{L,m}

heat taken in warms and changes phase(mixture)

s_{s} = s_{w}

Steam

T_{2,s} = T_{2,w}

the temperature of the mixture

Q_{w} = Q_{L,s} - Q_{s}

Steam is passed into (heat to warm of water)

m_{m} = m_{w} + m_{s}

Steam is passed into (mass of mixture)