Answer:
i)
ii)
ii)
Explanation:
As the initial and final states of the sample are the same, the ΔU of the sample is, for the three cases
since
i)Reversibly so
can be calculated by
and because of the first law of thermodynamics
ii)Irreversibly with
we can calculate by the law of ideal gases
then w can be calculated by
and
iii)a free expansion so
(there's no work at vaccum) and
Answer:
Due to the specific heat capacity of iron, 0.444 J/(g·°C), is more than the specific heat capacity for lead, 0.160 J/(g·°C)
Explanation:
The given parameters are;
The metals provided to melt the ice and their temperature includes;
One kg (1000 g) of iron;
Specific heat capacity = 0.444 J/(g·°C)
Temperature = 100°C
1 kg (1000 g) of lead
Specific heat capacity = 0.160 J/(g·°C)
Temperature = 100°C
Therefore, the heat provided to the ice of mass m, and latent heat of 334 J/g at 0°C by the metals are as follows;
For iron, we have;
ΔQ = Mass × Specific heat capacity × Temperature change
ΔQ = Heat obtained from the iron by the ice
ΔQ = 0.444 m × 1000 × (100 - 0) = 44400 J
Heat absorbed by the ice for melting, H = Heat obtained from the iron
∴ Heat absorbed by the ice for melting, H = Mass of ice × Latent heat of ice
H = Mass of ice × 334 J/g = 44400 J
Mass of ice melted by the iron = 44400 J/334 (J/g) ≈ 132.9 g
Mass of ice melted by the iron ≈ 132.9 g
For lead, we have;
ΔQ = Mass × Specific heat capacity × Temperature change
ΔQ = Heat obtained from the iron by the ice
ΔQ = 0.160 m × 1000 × (100 - 0) = 16000 J
Heat absorbed by the ice for melting, H = Heat obtained from the iron
∴ Heat absorbed by the ice for melting, H = Mass of ice × Latent heat of ice
H = Mass of ice × 334 J/g = 16000 J
Mass of ice melted by the lead = 16000 J/334 (J/g) ≈ 47.9 g
Mass of ice melted by the lead ≈ 47.9 g
Therefore, mass of ice melted by the iron, approximately 132.9 g, is more than mass of ice melted by the lead, approximately 47.9 g.