Fifty Grams Of Ice In Water At Eighty

Q. 50 g of ice at 0 °C is mixed with 50 g of water at 80 °C, final temperature of mixture will be expects A

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Q. 50 g $m_{i} = 50$ g of ice at $L_{i} = 80$ cal/g, $s_{i} = s_{w}$ 0 °C $T_{1,i} = 0$ °C is mixed with 50 g $m_{w} = 50$ g of water at $s_{w} = 1$ cal/g °C, $Q_{L,w} = 0$ J 80 °C $T_{1,w} = 80$ °C, final temperature of mixture $T_{2,i} = T_{2,w}$ , $Q_{in,i} = - Q_{in,w}$ will be

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The solution

Solved with Calorimetry

Given

$m_{i} = 50$ g
$T_{1,i} = 0$ °C
$m_{w} = 50$ g
$T_{1,w} = 80$ °C
$s_{w} = 1$ cal/g °C from “of water at”
$Q_{L,w} = 0$ J from “of water at”
$L_{i} = 80$ cal/g from “of ice at”
$s_{i} = s_{w}$ from “of ice at”
$T_{2,i} = T_{2,w}$ from “final temperature of mixture”
$Q_{in,i} = - Q_{in,w}$ from “final temperature of mixture”

Find $T_{2,w}$ .

Working

$m_{i} = 50$ g $= 0.05$ kg
$T_{1,i} = 0$ °C $= 273.15$ K
$m_{w} = 50$ g $= 0.05$ kg
$T_{1,w} = 80$ °C $= 353.15$ K
$s_{w} = 1$ cal/g °C $= 4186.8$ J kg⁻¹ K⁻¹
$L_{i} = 80$ cal/g $= 334944$ J/kg
heat changes a phase(ice) $Q_{L,i} = m_{i} \cdot L_{i}$
$Q_{L,i} = 0.05 \cdot 334944$
$Q_{L,i} = 16747.2$ J
of ice at $s_{i} = s_{w}$
$s_{i} = 4186.8$
$s_{i} = 4186.8$ J kg⁻¹ K⁻¹
heat taken in warms and changes phase(ice) $Q_{in,i} = Q_{i} + Q_{L,i}$
final temperature of mixture (heat taken in of ice) $Q_{in,i} = - Q_{in,w}$
heat taken in warms and changes phase(water) $Q_{in,w} = Q_{w} + Q_{L,w}$
heat warms a body(water) $Q_{w} = m_{w} \cdot s_{w} \cdot (T_{2,w} - T_{1,w})$
heat warms a body(ice) $Q_{i} = m_{i} \cdot s_{i} \cdot (T_{2,i} - T_{1,i})$
final temperature of mixture (temperature after of ice) $T_{2,i} = T_{2,w}$
$T_{2,w} = 0$ °C
$T_{2,w} = 273.15$ K $= 0$ °C

Conclusion

Each gram of water cooling 80 degrees gives the 80 cal a gram of ice needs to melt, and the masses are equal, so the ice just melts and the mixture stays at 0 °C.

The Ruby

1require_relative "../physics"23question :fifty_grams_of_ice_in_water_at_eighty do4  kind :scq5  describes "%{ice_mass} of ice at %{ice_temperature} is mixed with %{water_mass} of water at %{water_temperature}, " \6            "final temperature of mixture will be"7  given ice_mass: "50 g"8  given ice_temperature: "0 °C"9  given water_mass: "50 g"10  given water_temperature: "80 °C"1112  option "0 °C"13  option "40 °C"14  option "30 °C"15  option "4 °C"16end

1require_relative "../questions/fifty_grams_of_ice_in_water_at_eighty"2require_relative "../science/thermal_properties_of_matter/calorimetry"34solve :fifty_grams_of_ice_in_water_at_eighty, using: Calorimetry do5  objects :ice, :water67  take ice_mass: ice.mass, ice_temperature: ice.temperature_at_first8  take water_mass: water.mass, water_temperature: water.temperature_at_first910  deduce_from "of water at", water.specific_heat_capacity => "1 cal/g °C", water.heat_to_change_phase => "0 J"11  deduce_from "of ice at", ice.latent_heat => "80 cal/g", ice.specific_heat_capacity => water.specific_heat_capacity12  deduce_from "final temperature of mixture", ice.temperature_after => water.temperature_after,13                                              ice.heat_taken_in => -water.heat_taken_in1415  answer find_value(water.temperature_after, in: "°C")1617  conclusion "Each gram of water cooling 80 degrees gives the 80 cal a gram of ice needs to melt, and the masses " \18             "are equal, so the ice just melts and the mixture stays at 0 °C."19end

1require_relative "../solutions/fifty_grams_of_ice_in_water_at_eighty"23play_with :fifty_grams_of_ice_in_water_at_eighty do4  title "Fifty Grams Of Ice In Water At Eighty"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)

s_{i} = s_{w}

of ice at

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

final temperature of mixture (temperature after of ice)

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

final temperature of mixture (heat taken in of ice)