Melting Ice in Warm Water

Q. When 0.15 kg of ice at 0 °C is mixed with 0.30 kg of water at 50 °C in a container, the resulting temperature is 6.7 °C. Taking the specific heat capacity of water as 4186 J kg⁻¹ K⁻¹, what is the heat of fusion of ice, in J/kg? expects 3.34 × 10⁵ J/kg

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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?

The playground

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}

The solution

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 Ruby

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

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

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

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)