Physics to Go 1. Calculate the energy used, in joules, by each of the following: a) a \( 1500-\mathrm{W} \) hair dryer operating for 3 min b) a \( 1200-\mathrm{W} \) hair dryer operating for 4 min 2. If both situations described in Question 1 result in the same dryness of hair, which hair dryer is more efficient? 3. A 1200 -W hair dryer is used by several members of a family for a total of 30 min per day during a 30 -day month. How much electrical energy is consumed by the hair dryer during the month? Give your answer in: a) watt-hours b) kilowatt-hours 4. If the power company charges \( \$ 0.15 \) per \( \mathrm{kW} \cdot \mathrm{h} \) for electrical energy, what is the cost of using the hair dryer in Question 3 during the month? What is the cost for a year? 5. Not enough heat from the furnace reaches one bedroom in a home. The homeowner uses a portable 1350 -W electric heater 24 h per day to keep the bedroom warm during four cold winter months. At \( \$ 0.12 \) per kilowatt-hour, how much does it cost to operate the heater for the four months? (Assume two 30-day and two 31-day months.) 6. A portable CD player is rated at approximately 20 W and uses 4 AA batteries. a) Estimate the number of hours that you can listen to the music on a CD player before the batteries need replacing. b) Calculate the energy requirements of the CD player. c) Estimate the cost of 4 AA batteries. d) Calculate the cost per kilowatt-hour of a battery. e) Compare battery costs with the cost of electricity from the utilities (use approximately \( \$ 0.10 \) per kilowatt-hour).

Sure, let's work through each of these physics problems step by step. --- ## **1. Calculate the energy used, in joules, by each of the following:** ### **a) A \( 1500\,\mathrm{W} \) hair dryer operating for 3 minutes** **Given:** - Power (\( P \)) = 1500 W - Time (\( t \)) = 3 minutes = 180 seconds **Formula:** \[ \text{Energy} (E) = P \times t \] **Calculation:** \[ E = 1500\,\mathrm{W} \times 180\,\mathrm{s} = 270,000\,\mathrm{J} \] **Answer:** The hair dryer uses **270,000 joules** of energy. --- ### **b) A \( 1200\,\mathrm{W} \) hair dryer operating for 4 minutes** **Given:** - Power (\( P \)) = 1200 W - Time (\( t \)) = 4 minutes = 240 seconds **Formula:** \[ \text{Energy} (E) = P \times t \] **Calculation:** \[ E = 1200\,\mathrm{W} \times 240\,\mathrm{s} = 288,000\,\mathrm{J} \] **Answer:** The hair dryer uses **288,000 joules** of energy. --- ## **2. Efficiency Comparison of the Two Hair Dryers** **Given:** - Both hair dryers dry hair equally. - Energy used: - 1500 W: 270,000 J - 1200 W: 288,000 J **Analysis:** Efficiency is defined as the usefulness of output per unit of energy input. Since both dryers achieve the same result (dryness) but the 1500 W dryer uses less energy: **Conclusion:** The **1500 W hair dryer is more efficient**, as it accomplishes the same task using less energy. --- ## **3. Electrical Energy Consumption of a 1200 W Hair Dryer Over a Month** ### **a) In Watt-hours (Wh)** **Given:** - Power (\( P \)) = 1200 W - Usage per day (\( t \)) = 30 minutes = 0.5 hours - Number of days = 30 **Formula:** \[ \text{Energy} (E) = P \times t \times \text{days} \] **Calculation:** \[ E = 1200\,\mathrm{W} \times 0.5\,\mathrm{h} \times 30 = 18,000\,\mathrm{Wh} \] **Answer:** The hair dryer consumes **18,000 watt-hours** of electrical energy in the month. ### **b) In Kilowatt-hours (kWh)** **Conversion:** \[ 1\,\mathrm{kWh} = 1000\,\mathrm{Wh} \] **Calculation:** \[ E = \frac{18,000\,\mathrm{Wh}}{1000} = 18\,\mathrm{kWh} \] **Answer:** The energy consumption is **18 kilowatt-hours**. --- ## **4. Cost of Using the Hair Dryer** **Given:** - Energy consumed = 18 kWh (from Problem 3) - Cost per kWh = \$0.15 ### **Monthly Cost** **Formula:** \[ \text{Cost} = \text{Energy} \times \text{Rate} \] **Calculation:** \[ \text{Cost}_{\text{month}} = 18\,\mathrm{kWh} \times \$0.15/\mathrm{kWh} = \$2.70 \] ### **Annual Cost** **Given:** - Monthly cost = \$2.70 - Months in a year = 12 **Calculation:** \[ \text{Cost}_{\text{year}} = \$2.70 \times 12 = \$32.40 \] **Answers:** - **Monthly Cost:** \$2.70 - **Annual Cost:** \$32.40 --- ## **5. Operating Cost of a Portable Electric Heater Over Four Winter Months** **Given:** - Power (\( P \)) = 1350 W - Usage = 24 hours/day - Duration = Four months (two 30-day and two 31-day months) - Cost per kWh = \$0.12 **Calculations:** ### **Total Days:** \[ 2 \times 30 + 2 \times 31 = 60 + 62 = 122\,\text{days} \] ### **Total Hours:** \[ 24\,\text{hours/day} \times 122\,\text{days} = 2,928\,\text{hours} \] ### **Energy Consumed:** Convert power to kilowatts: \[ 1350\,\mathrm{W} = 1.35\,\mathrm{kW} \] \[ E = 1.35\,\mathrm{kW} \times 2,928\,\mathrm{h} = 3,952.8\,\mathrm{kWh} \] ### **Cost:** \[ \text{Cost} = 3,952.8\,\mathrm{kWh} \times \$0.12/\mathrm{kWh} = \$474.34 \] **Answer:** Operating the heater for four months costs **\$474.34**. --- ## **6. Analysis of a Portable CD Player Using 4 AA Batteries** ### **a) Estimated Listening Hours Before Batteries Need Replacing** **Given:** - Power (\( P \)) = 20 W - Number of AA batteries = 4 **Assumptions:** - Each AA battery has a capacity of approximately 2,500 mAh at 1.5 V. **Calculations:** 1. **Energy per Battery:** \[ E = V \times C = 1.5\,\mathrm{V} \times 2.5\,\mathrm{Ah} = 3.75\,\mathrm{Wh} \] 2. **Total Energy for 4 Batteries:** \[ E_{\text{total}} = 4 \times 3.75\,\mathrm{Wh} = 15\,\mathrm{Wh} \] 3. **Listening Time:** \[ t = \frac{E_{\text{total}}}{P} = \frac{15\,\mathrm{Wh}}{20\,\mathrm{W}} = 0.75\,\mathrm{hours} \] \[ 0.75\,\mathrm{hours} = 45\,\mathrm{minutes} \] **Answer:** You can listen to music for approximately **45 minutes** before needing to replace the batteries. --- ### **b) Calculate the Energy Requirements of the CD Player** **Given:** - Power (\( P \)) = 20 W - Usage time from part a) = 0.75 hours **Formula:** \[ \text{Energy} (E) = P \times t \] **Calculation:** \[ E = 20\,\mathrm{W} \times 0.75\,\mathrm{h} = 15\,\mathrm{Wh} \] **Answer:** The CD player requires **15 watt-hours** of energy. --- ### **c) Estimate the Cost of 4 AA Batteries** **Given:** - Cost per AA battery ≈ \$0.50 (prices may vary) **Calculation:** \[ \text{Total Cost} = 4 \times \$0.50 = \$2.00 \] **Answer:** 4 AA batteries cost approximately **\$2.00**. --- ### **d) Calculate the Cost per Kilowatt-hour of a Battery** **Given:** - Total energy from 4 batteries = 15 Wh - Total cost = \$2.00 **Conversion:** \[ 15\,\mathrm{Wh} = 0.015\,\mathrm{kWh} \] **Formula:** \[ \text{Cost per kWh} = \frac{\text{Total Cost}}{\text{Energy in kWh}} \] **Calculation:** \[ \text{Cost per kWh} = \frac{\$2.00}{0.015\,\mathrm{kWh}} = \$133.33/\mathrm{kWh} \] **Answer:** The cost per kilowatt-hour of the batteries is **\$133.33/kWh**. --- ### **e) Compare Battery Costs with the Cost of Electricity from Utilities** **Given:** - Cost per kWh from utilities ≈ \$0.10 **Comparison:** - **Battery Cost:** \$133.33/kWh - **Utility Cost:** \$0.10/kWh **Analysis:** \[ \$133.33/\mathrm{kWh} \gg \$0.10/\mathrm{kWh} \] **Conclusion:** Using batteries to supply energy is **significantly more expensive** than using electricity from the grid. Specifically, batteries cost over **1,300 times more per kilowatt-hour** than utility electricity. --- ## **Summary** - **Problem 1:** Calculated energy usage for hair dryers. - **Problem 2:** Determined the more efficient hair dryer based on energy consumption. - **Problem 3:** Computed monthly energy usage in Wh and kWh. - **Problem 4:** Calculated both monthly and annual costs of operating the hair dryer. - **Problem 5:** Evaluated the cost of operating an electric heater over four months. - **Problem 6:** Analyzed the energy usage and cost-effectiveness of using batteries versus utility power for a CD player. If you have any further questions or need additional explanations on any of these problems, feel free to ask!