Calculate the mass of dry ice that should be added to the water so that the dry ice completely sublimes away when the water reac
hes 16 ∘c. assume no heat loss to the surroundings.
1 answer:
There are a lot of missing information in this problem. I've found a similar problem which is shown in the attached picture. Let's just use the information there that we don't have here.
From the principle of conservation of energy:
Heat of dry ice + Heat of water = 0
Heat of dry ice = - Heat of water
(m of dry ice)ΔH = -(m of water)(Cp)(ΔT)
where Cp for water is 4.187 kJ/kg·°C
Hence,
(m of dry ice)(ΔH°g - ΔH°s) = -(Density*Volume)(Cp)(ΔT)
where the density of water is 1 kg/L and the molar mass of dry ice is 44 g/mol.
Then,
(m of dry ice)(-393.5 - -427.4 kJ/mol)(44 g/mol) = -(1 kg/L*12L)(4.187 kJ/kg·°C)(16 - 88 °C)
Solving for m of dry ice,
<em>Mass = 2.43 g of dry ice</em>
From the principle of conservation of energy:
Heat of dry ice + Heat of water = 0
Heat of dry ice = - Heat of water
(m of dry ice)ΔH = -(m of water)(Cp)(ΔT)
where Cp for water is 4.187 kJ/kg·°C
Hence,
(m of dry ice)(ΔH°g - ΔH°s) = -(Density*Volume)(Cp)(ΔT)
where the density of water is 1 kg/L and the molar mass of dry ice is 44 g/mol.
Then,
(m of dry ice)(-393.5 - -427.4 kJ/mol)(44 g/mol) = -(1 kg/L*12L)(4.187 kJ/kg·°C)(16 - 88 °C)
Solving for m of dry ice,
<em>Mass = 2.43 g of dry ice</em>
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Answer:
a)
b) entropy of the sistem equal to a), entropy of the universe grater than a).
Explanation:
a) The change of entropy for a reversible process:
The energy balance:
If the process is isothermical the U doesn't change:
The work:
If it is an ideal gas:
Solving:
Replacing:
Given that it's a compression: V2<V1 and ln(V2/V1)<0. So:
b) The entropy change of the sistem will be equal to the calculated in a), but the change of entropy of the universe will be 0 in a) (reversible process) and in b) has to be positive given that it is an irreversible process.