Ice to Steam — phryzby

solved Calorimetry

Q. Calculate the heat required $Q_{in,e} = Q_{in,i} + Q_{in,w}$ to convert $m_{w} = m_{i}$ 3 kg $m_{i} = 3$ kg of ice $T_{2,i} = 0$ °C, $T_{1,w} = 0$ °C at -12 °C $T_{1,i} = -12$ °C kept in a calorimeter to steam at 100 °C $T_{2,w} = 100$ °C at atmospheric pressure. Given specific heat capacity of ice = 2100 J kg⁻¹ K⁻¹ $s_{i} = 2100$ J kg⁻¹ K⁻¹, specific heat capacity of water = 4186 J kg⁻¹ K⁻¹ $s_{w} = 4186$ J kg⁻¹ K⁻¹, latent heat of fusion of ice = 3.35 × 10⁵ J/kg $L_{i} = 3.35 \times 10^{5}$ J/kg and latent heat of steam = 2.256 × 10⁶ J/kg $L_{w} = 2.256 \times 10^{6}$ J/kg. What is the heat required $Q_{in,e} = Q_{in,i} + Q_{in,w}$ , in joules?

Inputs

mass

m_{i}

3 kg

cold

T_{1,i}

-12 °C

boiling

T_{2,w}

100 °C

ice heat

s_{i}

2100 J kg⁻¹ K⁻¹

water heat

s_{w}

4186 J kg⁻¹ K⁻¹

fusion

L_{i}

3.35e+05 J/kg

steam

L_{w}

2.256e+06 J/kg

Outputs

heat

Q_{in,e}

How it's solved

Solved with Calorimetry

Given

$m_{i} = 3$ kg
$T_{1,i} = -12$ °C
$T_{2,w} = 100$ °C
$s_{i} = 2100$ J kg⁻¹ K⁻¹
$s_{w} = 4186$ J kg⁻¹ K⁻¹
$L_{i} = 3.35 \times 10^{5}$ J/kg
$L_{w} = 2.256 \times 10^{6}$ J/kg
$T_{2,i} = 0$ °C from “ice”
$T_{1,w} = 0$ °C from “ice”
$m_{w} = m_{i}$ from “convert”
$Q_{in,e} = Q_{in,i} + Q_{in,w}$ from “heat required”

Find $Q_{in,e}$ .

Working

$T_{1,i} = -12$ °C $= 261.15$ K
$T_{2,w} = 100$ °C $= 373.15$ K
$T_{2,i} = 0$ °C $= 273.15$ K
$T_{1,w} = 0$ °C $= 273.15$ K
heat warms a body(ice) $Q_{i} = m_{i} \cdot s_{i} \cdot (T_{2,i} - T_{1,i})$
$Q_{i} = 3 \cdot 2100 \cdot (273.15 - 261.15)$
$Q_{i} = 75600$ J
heat changes a phase(ice) $Q_{L,i} = m_{i} \cdot L_{i}$
$Q_{L,i} = 3 \cdot 335000$
$Q_{L,i} = 1.005 \times 10^{6}$ J
heat taken in warms and changes phase(ice) $Q_{in,i} = Q_{i} + Q_{L,i}$
$Q_{in,i} = 75600 + 1.005 \times 10^{6}$
$Q_{in,i} = 1.0806 \times 10^{6}$ J
convert $m_{w} = m_{i}$
$m_{w} = 3$
$m_{w} = 3$ kg
heat warms a body(water) $Q_{w} = m_{w} \cdot s_{w} \cdot (T_{2,w} - T_{1,w})$
$Q_{w} = 3 \cdot 4186 \cdot (373.15 - 273.15)$
$Q_{w} = 1.2558 \times 10^{6}$ J
heat changes a phase(water) $Q_{L,w} = m_{w} \cdot L_{w}$
$Q_{L,w} = 3 \cdot 2.256 \times 10^{6}$
$Q_{L,w} = 6.768 \times 10^{6}$ J
heat taken in warms and changes phase(water) $Q_{in,w} = Q_{w} + Q_{L,w}$
$Q_{in,w} = 1.2558 \times 10^{6} + 6.768 \times 10^{6}$
$Q_{in,w} = 8.0238 \times 10^{6}$ J
heat required $Q_{in,e} = Q_{in,i} + Q_{in,w}$
$Q_{in,e} = 1.0806 \times 10^{6} + 8.0238 \times 10^{6}$
$Q_{in,e} = 9.104 \times 10^{6}$ J
$Q_{in,e} \approx 9.1 \times 10^{6}$ Jto 2 significant figures

Conclusion

The ice warms to its melting point, melts, the water warms to its boiling point, and boils; boiling takes by far the most.

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 :ice_to_steam do4  kind :numerical5  describes "Calculate the heat required to convert %{mass} of ice at %{cold} kept in a calorimeter to steam at " \6            "%{boiling} at atmospheric pressure. Given specific heat capacity of ice = %{ice_heat}, specific heat " \7            "capacity of water = %{water_heat}, latent heat of fusion of ice = %{fusion} and latent heat of steam = " \8            "%{steam}. %{heat}, in joules?"9  given mass: "3 kg"10  given cold: "-12 °C"11  given boiling: "100 °C"12  given ice_heat: "2100 J kg-1 K-1"13  given water_heat: "4186 J kg-1 K-1"14  given fusion: "3.35e5 J/kg"15  given steam: "2.256e6 J/kg"16  asks heat: "What is the heat required"17end

solution.rb

1require_relative "../questions/ice_to_steam"2require_relative "../science/thermal_properties_of_matter/calorimetry"34solve :ice_to_steam, using: Calorimetry do5  objects :ice, :water, :everything67  take mass: ice.mass8  take cold: ice.temperature_at_first9  take boiling: water.temperature_after10  take ice_heat: ice.specific_heat_capacity11  take water_heat: water.specific_heat_capacity12  take fusion: ice.latent_heat13  take steam: water.latent_heat1415  deduce_from "convert", water.mass => ice.mass16  deduce_from "ice", ice.temperature_after => "0 °C", water.temperature_at_first => "0 °C"17  deduce_from "heat required", everything.heat_taken_in => ice.heat_taken_in + water.heat_taken_in1819  heat = everything.heat_taken_in20  answer heat:, in: "J", figures: 22122  conclusion "The ice warms to its melting point, melts, the water warms to its boiling point, and boils; " \23             "boiling takes by far the most."24end

playground.rb

1require_relative "../solutions/ice_to_steam"23play_with :ice_to_steam do4  title "Ice to Steam"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 ice	$Q_{i}$	J	—
mass of ice	$m_{i}$	kg	0 … ∞
specific heat capacity of ice	$s_{i}$	J kg⁻¹ K⁻¹	—
temperature after of ice	$T_{2,i}$	K	—
temperature at first of ice	$T_{1,i}$	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 everything	$Q_{e}$	J	—
mass of everything	$m_{e}$	kg	0 … ∞
specific heat capacity of everything	$s_{e}$	J kg⁻¹ K⁻¹	—
temperature after of everything	$T_{2,e}$	K	—
temperature at first of everything	$T_{1,e}$	K	—
heat to change phase of ground	$Q_{L,g}$	J	—
latent heat of ground	$L_{g}$	J/kg	—
heat to change phase of ice	$Q_{L,i}$	J	—
latent heat of ice	$L_{i}$	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 everything	$Q_{L,e}$	J	—
latent heat of everything	$L_{e}$	J/kg	—
heat taken in of ground	$Q_{in,g}$	J	—
heat taken in of ice	$Q_{in,i}$	J	—
heat taken in of water	$Q_{in,w}$	J	—
heat taken in of everything	$Q_{in,e}$	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_{i} = m_{i} \cdot s_{i} \cdot (T_{2,i} - T_{1,i})

heat warms a body(ice)

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

heat warms a body(water)

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

heat warms a body(everything)

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

heat changes a phase(ground)

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

heat changes a phase(ice)

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

heat changes a phase(water)

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

heat changes a phase(everything)

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

heat taken in warms and changes phase(ground)

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

heat taken in warms and changes phase(ice)

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

heat taken in warms and changes phase(water)

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

heat taken in warms and changes phase(everything)

m_{w} = m_{i}

convert

Q_{in,e} = Q_{in,i} + Q_{in,w}

heat required