deremine (a) the work done and (b) the change in internal energy of 1.00 kg of water when it is all boiled to steam at 100c. Assume a constant pressure of 1.00 atm and heat of vaporisation = 2260000 J.
To determine (a) the work done and (b) the change in internal energy of 1.00 kg of water when it is boiled to steam at 100°C, we can use the first law of thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat added (Q) minus the work done (W), given as:
ΔU = Q - W
To calculate the work done, we use the formula:
W = P * ΔV
where P is the constant pressure and ΔV is the change in volume. Since the pressure is constant and the water is boiled to steam, we can assume that the volume change is the same as the liquid water's specific volume.
1. Determine the work done (W):
First, we need to calculate the change in volume. The specific volume of water at 100°C and 1 atm pressure is approximately 1.044 x 10^-3 m^3/kg.
ΔV = specific volume of steam - specific volume of water
= (1.044 x 10^-3 m^3/kg) - (1.044 x 10^-3 m^3/kg) (since steam and boiling water have approximately the same specific volume at this temperature)
ΔV = 0 m^3/kg (negligible)
Therefore, the work done (W) is equal to zero since ΔV is zero.
W = 0
2. Determine the change in internal energy (ΔU):
We can calculate the change in internal energy (ΔU) using the formula:
ΔU = Q - W
The heat added (Q) can be calculated using the given heat of vaporization value:
Q = mass * heat of vaporization
Where mass is the mass of the water, which is given as 1.00 kg.
Q = (1.00 kg) * (2,260,000 J/kg)
Q = 2,260,000 J
Now we can substitute the values into the formula:
ΔU = Q - W
= 2,260,000 J - 0 J (since W is zero)
= 2,260,000 J
Therefore, (a) the work done is 0 J and (b) the change in internal energy of 1.00 kg of water boiled to steam at 100°C is 2,260,000 J.
To determine the work done and the change in internal energy when 1.00 kg of water is boiled to steam at 100°C, we can use the first law of thermodynamics. The first law states that the change in internal energy (ΔU) of a system is equal to the heat (q) added to the system minus the work (W) done by the system:
ΔU = q - W
(a) To calculate the work done, we need to determine the volume change of water during the phase transition from liquid to gas. At constant pressure, the work done is given by:
W = P ΔV
Where P is the constant pressure and ΔV is the change in volume.
When water is boiled, it changes phase from a liquid to a gas (steam), and this phase transition occurs at a constant temperature of 100°C. During this process, the volume increases significantly. However, since we're assuming a constant pressure of 1.00 atm, we can use the ideal gas law to estimate the change in volume.
The ideal gas law is given by:
PV = nRT
Where P is pressure, V is volume, n is the number of moles of gas, R is the ideal gas constant, and T is temperature in Kelvin.
Since we are assuming 1.00 kg of water, we can determine the number of moles of water using its molecular weight (18.015 g/mol):
n = m/M
Where m is the mass of water and M is the molar mass of water.
n = 1.00 kg / 18.015 g/mol = 0.0555 mol
To calculate the initial volume of water, we need to know its density. The density of water is approximately 1000 kg/m³.
ρ = m/V
Where ρ is density, m is mass, and V is volume.
V = m / ρ = 1.00 kg / 1000 kg/m³ = 0.001 m³
At the boiling point (100°C), water starts to vaporize and the volume increases. The final volume of steam can be calculated using the relationship between the initial and final temperatures and volumes:
(V₁ / T₁) = (V₂ / T₂)
Where V₁ and V₂ are the initial and final volumes, and T₁ and T₂ are the initial and final temperatures in Kelvin.
We have the initial volume (V₁ = 0.001 m³) and temperature (T₁ = 100°C = 373 K). We are looking for the final volume (V₂) at the same pressure of 1.00 atm and temperature of (T₂ = 100°C = 373 K).
Solving for V₂:
V₂ = (V₁ * T₂) / T₁ = (0.001 m³ * 373 K) / 373 K = 0.001 m³
The change in volume (ΔV) is then:
ΔV = V₂ - V₁ = 0.001 m³ - 0.001 m³ = 0 m³
Since there is no change in volume (ΔV = 0), the work done (W) is:
W = P ΔV = 1.00 atm * 0 m³ = 0 J
Therefore, the work done when 1.00 kg of water is boiled to steam at 100°C is 0 J.
(b) To calculate the change in internal energy (ΔU), we need to know the heat (q) added to the system.
The heat required to vaporize water is known as the heat of vaporization. In this case, the heat of vaporization is given as 2,260,000 J/kg.
q = m * ΔH
Where ΔH is the heat of vaporization.
q = 1.00 kg * 2,260,000 J/kg = 2,260,000 J
Now, we can calculate the change in internal energy using the first law of thermodynamics:
ΔU = q - W = 2,260,000 J - 0 J = 2,260,000 J
Therefore, the change in internal energy of 1.00 kg of water when it is boiled to steam at 100°C is 2,260,000 J.