Melting Ice in Warm Water

solved Calorimetry

Q. When 0.15 kg $m_{i} = 0.15$ kg of ice at 0 °C $T_{1,i} = 0$ °C is mixed with 0.30 kg $m_{w} = 0.30$ kg of water at $Q_{L,w} = 0$ J 50 °C $T_{1,w} = 50$ °C in a container, the resulting temperature $T_{2,i} = T_{2,w}$ , $s_{i} = s_{w}$ , $Q_{in,i} = - Q_{in,w}$ is 6.7 °C $T_{2,w} = 6.7$ °C. Taking the specific heat capacity of water as 4186 J kg⁻¹ K⁻¹ $s_{w} = 4186$ J kg⁻¹ K⁻¹, what is the heat of fusion of ice, in J/kg?

Inputs

ice mass

m_{i}

0.15 kg

water mass

m_{w}

0.3 kg

water temperature

T_{1,w}

50 °C

mixed

T_{2,w}

6.7 °C

water heat

s_{w}

4186 J kg⁻¹ K⁻¹

Outputs

fusion

L_{i}

How it's solved

Solved with Calorimetry

Given

$m_{i} = 0.15$ kg
$T_{1,i} = 0$ °C
$m_{w} = 0.30$ kg
$T_{1,w} = 50$ °C
$T_{2,w} = 6.7$ °C
$s_{w} = 4186$ J kg⁻¹ K⁻¹
$Q_{L,w} = 0$ J from “of water at”
$T_{2,i} = T_{2,w}$ from “the resulting temperature”
$s_{i} = s_{w}$ from “the resulting temperature”
$Q_{in,i} = - Q_{in,w}$ from “the resulting temperature”

Find $L_{i}$ .

Working

$T_{1,i} = 0$ °C $= 273.15$ K
$T_{1,w} = 50$ °C $= 323.15$ K
$T_{2,w} = 6.7$ °C $= 279.85$ K
heat warms a body(water) $Q_{w} = m_{w} \cdot s_{w} \cdot (T_{2,w} - T_{1,w})$
$Q_{w} = 0.3 \cdot 4186 \cdot (279.85 - 323.15)$
$Q_{w} = −54376.14$ J
heat taken in warms and changes phase(water) $Q_{in,w} = Q_{w} + Q_{L,w}$
$Q_{in,w} = (−54376.14) + 0$
$Q_{in,w} = −54376.14$ J
the resulting temperature (heat taken in of ice) $Q_{in,i} = - Q_{in,w}$
$Q_{in,i} = - (−54376.14)$
$Q_{in,i} = 54376.14$ J
the resulting temperature (specific heat capacity of ice) $s_{i} = s_{w}$
$s_{i} = 4186$
$s_{i} = 4186$ J kg⁻¹ K⁻¹
the resulting temperature (temperature after of ice) $T_{2,i} = T_{2,w}$
$T_{2,i} = 279.85$
$T_{2,i} = 279.85$ K
heat warms a body(ice) $Q_{i} = m_{i} \cdot s_{i} \cdot (T_{2,i} - T_{1,i})$
$Q_{i} = 0.15 \cdot 4186 \cdot (279.85 - 273.15)$
$Q_{i} = 4206.93$ J
heat taken in warms and changes phase(ice) $Q_{in,i} = Q_{i} + Q_{L,i}$
$54376.14 = 4206.93 + Q_{L,i}$
$Q_{L,i} = 50169.21$ J
heat changes a phase(ice) $Q_{L,i} = m_{i} \cdot L_{i}$
$50169.21 = 0.15 \cdot L_{i}$
$L_{i} = 334500$ J/kg
$L_{i} \approx 3.34 \times 10^{5}$ J/kgto 3 significant figures

Conclusion

The warm water gives up heat that first melts the ice and then warms the melted ice to the mixture's temperature; what is left over after the warming is the heat of fusion.

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 :melting_ice_in_warm_water do4  kind :numerical5  describes "When %{ice_mass} of ice at %{ice_temperature} is mixed with %{water_mass} of water at " \6            "%{water_temperature} in a container, the resulting temperature is %{mixed}. Taking the specific heat " \7            "capacity of water as %{water_heat}, %{fusion}, in J/kg?"8  given ice_mass: "0.15 kg"9  given ice_temperature: "0 °C"10  given water_mass: "0.30 kg"11  given water_temperature: "50 °C"12  given mixed: "6.7 °C"13  given water_heat: "4186 J kg-1 K-1"14  asks fusion: "what is the heat of fusion of ice"15end

solution.rb

1require_relative "../questions/melting_ice_in_warm_water"2require_relative "../science/thermal_properties_of_matter/calorimetry"34solve :melting_ice_in_warm_water, using: Calorimetry do5  objects :ice, :water67  take ice_mass: ice.mass8  take ice_temperature: ice.temperature_at_first9  take water_mass: water.mass10  take water_temperature: water.temperature_at_first11  take mixed: water.temperature_after12  take water_heat: water.specific_heat_capacity1314  deduce_from "the resulting temperature", ice.temperature_after => water.temperature_after,15                                           ice.specific_heat_capacity => water.specific_heat_capacity,16                                           ice.heat_taken_in => -water.heat_taken_in17  deduce_from "of water at", water.heat_to_change_phase => "0 J"1819  fusion = ice.latent_heat20  answer fusion:, in: "J/kg", figures: 32122  conclusion "The warm water gives up heat that first melts the ice and then warms the melted ice to the " \23             "mixture's temperature; what is left over after the warming is the heat of fusion."24end

playground.rb

1require_relative "../solutions/melting_ice_in_warm_water"23play_with :melting_ice_in_warm_water do4  title "Melting Ice in Warm Water"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 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 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	—

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_{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_{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)

T_{2,i} = T_{2,w}

the resulting temperature (temperature after of ice)

s_{i} = s_{w}

the resulting temperature (specific heat capacity of ice)

Q_{in,i} = - Q_{in,w}

the resulting temperature (heat taken in of ice)