You use a \( 1500-\mathrm{W} \) hair dryer for 5 min every morning to dry your hair. a) How much electrical energy are you changing to heat every day? b) If you could transfer all of that energy to heating up water for a shower, how much water could you heat from \( 20^{\circ} \mathrm{C} \) to \( 45^{\circ} \mathrm{C} \) ? c) If one kilogram of water is about 0.26 gal (gallons), how many gallons of water does that represent? Is that enough water to take a shower? How much water do you think you use in a typical shower in your own home? d) Make a statement in your log about which use of this amount of energy would make most sense in your appliance package - using the hairdryer to

Sure, let's tackle each part of the problem step by step. --- ### **a) Calculating Daily Electrical Energy Consumption** **Given:** - Power of hair dryer, \( P = 1500 \, \text{W} \) - Usage time, \( t = 5 \, \text{minutes} \) **Calculations:** 1. **Convert Time to Seconds:** \[ t = 5 \, \text{minutes} \times 60 \, \frac{\text{seconds}}{\text{minute}} = 300 \, \text{seconds} \] 2. **Calculate Energy in Joules:** \[ E = P \times t = 1500 \, \text{W} \times 300 \, \text{s} = 450{,}000 \, \text{J} \] 3. **Alternatively, Convert Energy to Kilowatt-hours (kWh):** \[ \text{Power in kW} = \frac{1500 \, \text{W}}{1000} = 1.5 \, \text{kW} \] \[ t = \frac{5}{60} \, \text{hours} \approx 0.0833 \, \text{hours} \] \[ E = 1.5 \, \text{kW} \times 0.0833 \, \text{hours} \approx 0.125 \, \text{kWh} \] **Answer:** You are converting **450,000 joules (or approximately 0.125 kilowatt-hours)** of electrical energy to heat every day. --- ### **b) Heating Water with the Same Energy** **Given:** - Energy available, \( E = 450{,}000 \, \text{J} \) - Initial water temperature, \( T_{\text{initial}} = 20^\circ \text{C} \) - Final water temperature, \( T_{\text{final}} = 45^\circ \text{C} \) - Specific heat capacity of water, \( c = 4186 \, \frac{\text{J}}{\text{kg} \cdot \text{°C}} \) - Temperature change, \( \Delta T = 25^\circ \text{C} \) **Calculations:** 1. **Use the Specific Heat Formula:** \[ Q = m \times c \times \Delta T \] Solving for mass \( m \): \[ m = \frac{Q}{c \times \Delta T} = \frac{450{,}000 \, \text{J}}{4186 \, \frac{\text{J}}{\text{kg} \cdot \text{°C}} \times 25 \, \text{°C}} \approx 4.3 \, \text{kg} \] **Answer:** You could heat approximately **4.3 kilograms** of water from \( 20^\circ \text{C} \) to \( 45^\circ \text{C} \) with the daily energy used by the hair dryer. --- ### **c) Converting Kilograms of Water to Gallons and Assessing Shower Sufficiency** **Given:** - Mass of water, \( m = 4.3 \, \text{kg} \) - Conversion factor, \( 1 \, \text{kg} \approx 0.26 \, \text{gallons} \) **Calculations:** 1. **Convert Kilograms to Gallons:** \[ \text{Volume} = 4.3 \, \text{kg} \times 0.26 \, \frac{\text{gallons}}{\text{kg}} \approx 1.12 \, \text{gallons} \] 2. **Assessing Shower Sufficiency:** - **Typical Shower Usage:** An average shower uses about **10 to 25 gallons** of water, depending on duration and showerhead efficiency. - **Comparison:** **1.12 gallons** is significantly less than the water required for a standard shower. **Answer:** The energy could heat approximately **1.12 gallons** of water, which **is not enough** for a typical shower. Most showers use between **10 to 25 gallons** of water. --- ### **d) Statement on Energy Use in Appliance Package** **Statement:** *Considering the daily energy consumption of the hair dryer and its potential alternative use for heating water, it becomes evident that using the hair dryer to dry hair is a more practical application of this energy. Heating just over one gallon of water is insufficient for essential needs like taking a shower, which typically requires much more water. Therefore, integrating a hair dryer for personal grooming within the appliance package makes more sense, as it effectively utilizes the energy without compromising essential water heating requirements.*