Calculate the change in internal energy of 2kg of water at 90 degree Celsius when it is changed to

3.30m3
of steam at
100oC
. The whole process occurs at atmospheric pressure. The latent heat of vaporization of water is
2.26×106J/kg

ΔQ=ΔU+W

ΔU = ΔQ – W =
=c•m•ΔT + r•m – p•ΔV=
= c•m•ΔT + r•m – p{V- (m/ρ)} =
=4190•2•10 +2.26•10⁶•2 – 101325(3.3 – 2/1000) = 4515630 J

45.72mJ

Answer is du= mc(teta) + ml - pdv. Solving it gives you 4269830.2J

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To calculate the change in internal energy of water when it is changed to steam, we need to consider two parts of the process: the increase in temperature from 90°C to 100°C and the conversion of water to steam at 100°C.

Let's break down the steps to find the change in internal energy:

Step 1: Calculate the change in internal energy due to temperature increase.
The specific heat capacity of water, c, is approximately 4,186 J/(kg·°C). The formula to calculate the change in internal energy due to a temperature increase is:

ΔQ = mcΔT

Where:
ΔQ is the change in internal energy,
m is the mass of water,
c is the specific heat capacity of water,
ΔT is the change in temperature.

Given values:
m = 2 kg
c = 4186 J/(kg·°C)
ΔT = 100°C - 90°C = 10°C

Plug in these values:

ΔQ = (2 kg)(4186 J/(kg·°C))(10°C)
ΔQ = 83720 J

Step 2: Calculate the change in internal energy due to the phase change (water to steam).
The latent heat of vaporization, L, of water is given as 2.26×10^6 J/kg. The formula to calculate the change in internal energy due to a phase change is:

ΔQ = mL

Where:
ΔQ is the change in internal energy,
m is the mass of water,
L is the latent heat of vaporization of water.

Given values:
m = 2 kg
L = 2.26×10^6 J/kg

Plug in these values:

ΔQ = (2 kg)(2.26×10^6 J/kg)
ΔQ = 4.52×10^6 J

Step 3: Calculate the total change in internal energy.
To find the total change in internal energy, we need to sum up the changes in internal energy due to temperature increase and phase change:

ΔQ_total = ΔQ_temperature + ΔQ_phase_change

ΔQ_total = 83720 J + 4.52×10^6 J
ΔQ_total = 4.52×10^6 J + 83720 J
ΔQ_total = 4.61×10^6 J

The change in internal energy of 2 kg of water when it is changed to 3.30 m^3 of steam is approximately 4.61×10^6 J.