Problem — Morning shower

Problem

Estimate the heat needed to heat the water for a $5$-minute shower, from $15\,^\circ\text{C}$ to $40\,^\circ\text{C}$, with a flow rate of $9\;\text{L/min}$. Compare with the energy of an AA battery ($\sim 10\;\text{kJ}$).

Solution

Mass of water. In $5$ minutes at $9\;\text{L/min}$ the flow is

$V = 5\;\text{min} \times 9\;\text{L/min} = 45\;\text{L} \;\Rightarrow\; m = 45\;\text{kg}$

(recalling $1\;\text{L}$ of water $\approx 1\;\text{kg}$).

Heat needed to heat it. With the specific heat of water $c = 4186\;\text{J/(kg·K)}$ and a temperature rise $\Delta T = 40 - 15 = 25\;\text{K}$:

$Q = m\,c\,\Delta T = 45 \times 4186 \times 25 \approx \ev{4{,}7\;\text{MJ}}$

Comparison with the AA battery. An AA battery stores $\sim 10\;\text{kJ}$:

$N = \frac{Q}{E_\text{battery}} = \frac{4{,}7\times10^6\;\text{J}}{10^4\;\text{J}} \approx \ev{470\ \text{AA batteries}}$

Nearly $500$ AA batteries would be needed for a single shower: this makes clear just how much energy is hidden in simply heating water, and why the water heater is among the largest domestic energy consumers. The thermal energy for “mundane” uses far exceeds the electrochemical energy of small devices.

Collegamenti

Argomenti: Termodinamica Concetti: Primo principio della termodinamica · Energia interna Competenze: Stime di Fermi · Conservazione dell’energia