Copper Block On Ice

Part of eval backend edge, agent version bdd945f

solved Calorimetry

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?

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}

How it's solved

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 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

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

solution.rb

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

playground.rb

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

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 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