Ice to Steam — phryzby

Q. Calculate the heat required to convert 3 kg of ice at -12 °C kept in a calorimeter to steam at 100 °C at atmospheric pressure. Given specific heat capacity of ice = 2100 J kg⁻¹ K⁻¹, specific heat capacity of water = 4186 J kg⁻¹ K⁻¹, latent heat of fusion of ice = 3.35 × 10⁵ J/kg and latent heat of steam = 2.256 × 10⁶ J/kg. What is the heat required, in joules? expects 9.1 × 10⁶ J

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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?

The playground

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}

The solution

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 Ruby

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

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

1require_relative "../solutions/ice_to_steam"23play_with :ice_to_steam do4  title "Ice to Steam"5end

The laws

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