Copper Block On Ice

Part of eval backend edge, agent version bdd945f

Q. A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500 °C and then placed on a large ice block. Specific heat of copper = 0.39 J g⁻¹ K⁻¹; heat of fusion of water = 335 J g⁻¹. What is the maximum amount of ice that can melt, in kg? expects 1.5 kg

7 checks, 1 agreed

draft 7: agreed

Latest draft

question.rb1 question :copper_block_on_ice do2   kind :numerical3   describes "A copper block of %{block_mass} is heated in a furnace to a temperature of %{hot} and then placed on a large ice block. Specific heat of copper = %{copper_heat}; heat of fusion of water = %{fusion}. %{ice_melted}, in kg?"4   given block_mass: "2.5 kg"5   given hot: "500 °C"6   given copper_heat: "0.39 J g-1 K-1"7   given fusion: "335 J g-1"8   asks ice_melted: "What is the maximum amount of ice that can melt"9 end

solution.rb1 solve :copper_block_on_ice, using: Calorimetry do2   objects :block, :ice3 4   take block_mass: block.mass5   take hot: block.temperature_at_first6   take copper_heat: block.specific_heat_capacity7   take fusion: ice.latent_heat8 9   deduce_from "placed on a large ice block", ice.temperature_at_first => "0 °C"10   deduce_from "maximum amount of ice that can melt", block.temperature_after => "0 °C", block.heat_to_warm => -ice.heat_to_change_phase11 12   ice_melted = ice.mass13   answer ice_melted:, in: "kg", figures: 214 15   conclusion "The copper block cools to 0 °C, giving up heat that melts ice at 0 °C; the heat lost by the block equals the heat gained by the ice to change phase."16 end

round 7

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Q. A copper block of 2.5 kg $m_{b} = 2.5$ kg is heated in a furnace to a temperature of 500 °C $T_{1,b} = 500$ °C and then placed on a large ice block $T_{1,i} = 0$ °C. Specific heat of copper = 0.39 J g⁻¹ K⁻¹ $s_{b} = 0.39$ J g⁻¹ K⁻¹; heat of fusion of water = 335 J g⁻¹ $L_{i} = 335$ J g⁻¹. What is the maximum amount of ice that can melt $T_{2,b} = 0$ °C, $Q_{b} = - Q_{L,i}$ , in kg?

The playground

Inputs

block mass

m_{b}

2.5 kg

hot

T_{1,b}

500 °C

copper heat

s_{b}

0.39 J g⁻¹ K⁻¹

fusion

L_{i}

335 J g⁻¹

Outputs

ice melted

m_{i}

The solution

Solved with Calorimetry

Given

$m_{b} = 2.5$ kg
$T_{1,b} = 500$ °C
$s_{b} = 0.39$ J g⁻¹ K⁻¹
$L_{i} = 335$ J g⁻¹
$T_{1,i} = 0$ °C from “placed on a large ice block”
$T_{2,b} = 0$ °C from “maximum amount of ice that can melt”
$Q_{b} = - Q_{L,i}$ from “maximum amount of ice that can melt”

Find $m_{i}$ .

Working

$T_{1,b} = 500$ °C $= 773.15$ K
$s_{b} = 0.39$ J g⁻¹ K⁻¹ $= 390$ J kg⁻¹ K⁻¹
$L_{i} = 335$ J g⁻¹ $= 335000$ J/kg
$T_{1,i} = 0$ °C $= 273.15$ K
$T_{2,b} = 0$ °C $= 273.15$ K
heat warms a body(block) $Q_{b} = m_{b} \cdot s_{b} \cdot (T_{2,b} - T_{1,b})$
$Q_{b} = 2.5 \cdot 390 \cdot (273.15 - 773.15)$
$Q_{b} = −487500$ J
maximum amount of ice that can melt $Q_{b} = - Q_{L,i}$
$(−487500) = - Q_{L,i}$
$Q_{L,i} = 487500$ J
heat changes a phase(ice) $Q_{L,i} = m_{i} \cdot L_{i}$
$487500 = m_{i} \cdot 335000$
$m_{i} = 1.455$ kg
$m_{i} \approx 1.5$ kgto 2 significant figures

Conclusion

The copper block cools to 0 °C, giving up heat that melts ice at 0 °C; the heat lost by the block equals the heat gained by the ice to change phase.

The Ruby

1question :copper_block_on_ice do2  kind :numerical3  describes "A copper block of %{block_mass} is heated in a furnace to a temperature of %{hot} and then placed on a large ice block. Specific heat of copper = %{copper_heat}; heat of fusion of water = %{fusion}. %{ice_melted}, in kg?"4  given block_mass: "2.5 kg"5  given hot: "500 °C"6  given copper_heat: "0.39 J g-1 K-1"7  given fusion: "335 J g-1"8  asks ice_melted: "What is the maximum amount of ice that can melt"9end

1solve :copper_block_on_ice, using: Calorimetry do2  objects :block, :ice34  take block_mass: block.mass5  take hot: block.temperature_at_first6  take copper_heat: block.specific_heat_capacity7  take fusion: ice.latent_heat89  deduce_from "placed on a large ice block", ice.temperature_at_first => "0 °C"10  deduce_from "maximum amount of ice that can melt", block.temperature_after => "0 °C", block.heat_to_warm => -ice.heat_to_change_phase1112  ice_melted = ice.mass13  answer ice_melted:, in: "kg", figures: 21415  conclusion "The copper block cools to 0 °C, giving up heat that melts ice at 0 °C; the heat lost by the block equals the heat gained by the ice to change phase."16end

1play_with :copper_block_on_ice do2  title "Copper Block On Ice"3end

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 block	$Q_{b}$	J	—
mass of block	$m_{b}$	kg	0 … ∞
specific heat capacity of block	$s_{b}$	J kg⁻¹ K⁻¹	—
temperature after of block	$T_{2,b}$	K	—
temperature at first of block	$T_{1,b}$	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 change phase of ground	$Q_{L,g}$	J	—
latent heat of ground	$L_{g}$	J/kg	—
heat to change phase of block	$Q_{L,b}$	J	—
latent heat of block	$L_{b}$	J/kg	—
heat to change phase of ice	$Q_{L,i}$	J	—
latent heat of ice	$L_{i}$	J/kg	—
heat taken in of ground	$Q_{in,g}$	J	—
heat taken in of block	$Q_{in,b}$	J	—
heat taken in of ice	$Q_{in,i}$	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_{b} = m_{b} \cdot s_{b} \cdot (T_{2,b} - T_{1,b})

heat warms a body(block)

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

heat warms a body(ice)

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

heat changes a phase(ground)

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

heat changes a phase(block)

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

heat changes a phase(ice)

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

heat taken in warms and changes phase(ground)

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

heat taken in warms and changes phase(block)

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

heat taken in warms and changes phase(ice)

Q_{b} = - Q_{L,i}

maximum amount of ice that can melt