Solar houses use a variety of energy storage devices to retain the heat absorbed during the day so
that it can be released during the night. Suppose that you were to use a device of this kind to
produce steam at $100 ^ { \circ } \mathrm { C }$ during the day, and then allow the steam to cool to $0 ^ { \circ } \mathrm { C }$ and freeze during
the night. How many kilograms of water would be needed to store 20.0 kWh of energy in this
way? The latent heat of vaporization of water is $22.6 \times 10 ^ { 5 } \mathrm {~J} / \mathrm { kg }$ the latent heat of fusion of water is $33.5 \times 10 ^ { 4 } \mathrm {~J} / \mathrm { kg }$ and the specific heat capacity of water is $4186 \mathrm {~J} / \mathrm { kg } \cdot \mathrm { K }$

Question

Solar houses use a variety of energy storage devices to retain the heat absorbed during the day so
that it can be released during the night. Suppose that you were to use a device of this kind to
produce steam at $100 ^ { \circ } \mathrm { C }$ during the day, and then allow the steam to cool to $0 ^ { \circ } \mathrm { C }$ and freeze during
the night. How many kilograms of water would be needed to store  20.0 kWh of energy in this
way? The latent heat of vaporization of water is $22.6 \times 10 ^ { 5 } \mathrm {~J} / \mathrm { kg }$ the latent heat of fusion of water is $33.5 \times 10 ^ { 4 } \mathrm {~J} / \mathrm { kg }$ and the specific heat capacity of water is $4186 \mathrm {~J} / \mathrm { kg } \cdot \mathrm { K }$

Quizplus · Accepted Answer

To find the mass of water needed to store 20.0 kWh of energy, we first convert the energy from kWh to joules: $20.0 \, \text{kWh} = 20.0 \times 3.6 \times 10^6 \, \text{J} = 72.0 \times 10^6 \, \text{J}$.The total energy stored will be the sum of the energy from the phase change of vaporization, the cooling of steam from $100^\circ C$ to $0^\circ C$, and the phase change of fusion.1. Energy from vaporization: $E_v = m \times L_v$2. Energy from cooling: $E_c = m \times c \times \Delta T$3. Energy from fusion: $E_f = m \times L_f$Given:- $L_v = 22.6 \times 10^5 \, \text{J/kg}$- $L_f = 33.5 \times 10^4 \, \text{J/kg}$- $c = 4186 \, \text{J/kg} \cdot \text{K}$- $\Delta T = 100 \, \text{K}$ (from $100^\circ C$ to $0^\circ C$)The total energy $E_{total}$ is the sum of $E_v$, $E_c$, and $E_f$, and we set this equal to the energy we want to store, $72.0 \times 10^6 \, \text{J}$:$$E_{total} = m \times L_v + m \times c \times \Delta T + m \times L_f = 72.0 \times 10^6 \, \text{J}$$Substituting the given values:$$m \times (22.6 \times 10^5 + 4186 \times 100 + 33.5 \times 10^4) = 72.0 \times 10^6$$$$m \times (22.6 \times 10^5 + 418600 + 33.5 \times 10^4) = 72.0 \times 10^6$$$$m \times (2.26 \times 10^6 + 4.186 \times 10^5 + 3.35 \times 10^5) = 72.0 \times 10^6$$$$m \times (3.0036 \times 10^6) = 72.0 \times 10^6$$$$m = \frac{72.0 \times 10^6}{3.0036 \times 10^6} \approx 23.9 \, \text{kg}$$Therefore, approximately 23.9 kg of water would be needed to store 20.0 kWh of energy in this way.

Solar Houses Use a Variety of Energy Storage Devices to Retain