Question

Calculate the change in internal energy of 2kg of water at
90oC90oC
when it is changed to
3.30m33.30m3
of steam at
100oC100oC
. The whole process occurs at atmospheric pressure. The latent heat of vaporization of water is
2.26×106J/kg2.26×106J/kg
.
4.27 MJ
3.43 kJ
45.72 mJ
543.63 J

Answer 1

Answer on Question #58707, Physics / Molecular Physics | Thermodynamics |

Calculate the change in internal energy of $2\mathrm{kg}$ of water at $90{}^{\circ}\mathrm{C}$ when it is changed to $3.30\mathrm{m}^{3}$ of steam at $100{}^{\circ}\mathrm{C}$. The whole process occurs at atmospheric pressure. The latent heat of vaporization of water is $2.26\times 10^{6}\mathrm{J / kg}$.

4.27 MJ

3.43 kJ

45.72 mJ

543.63 J

Solution:

The amount of heat received by water is equal to the sum of the change in internal energy of water and the work on the steam:

$$Q = \Delta U + W$$

where

$$Q = Q _ {h e a t} + Q _ {v a p o r}$$ $$Q _ {h e a t} = C m \Delta T$$ $$Q _ {v a p o r} = r m$$

where $r$ is the latent heat of vaporization of water.

The work is

$$W = p \Delta V = p \left(V _ {s t e a m} - \frac {m}{\rho}\right)$$

where $\rho = 1000\mathrm{kg / m^3}$ is the density of water.

The change of the internal energy is

$$\begin{array}{l} \Delta U = Q - W = C m \Delta T + r m - p \Delta V = \\ = 4 2 0 0 \cdot 2 \cdot (1 0 0 - 9 0) + 2. 2 6 \cdot 1 0 ^ {6} \cdot 2 - 1 \cdot 1 0 ^ {5} \left(3. 3 0 - \frac {2}{1 0 0 0}\right) = 4 2 7 4 2 0 0 J = 4. 2 7 M J \\ \end{array}$$

Answer: 4.27 MJ.

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