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Chapter 7: Problem 125

The best solar panels currently available are about \(15 \%\) efficient in converting sunlight to electricity. A typical home will use about \(40 .\) kWh of electricity per day \((1 \mathrm{kWh}=1\) kilowatt hour; \(1 \mathrm{kW}=1000 \mathrm{J} / \mathrm{s}\) ). Assuming 8.0 hours of useful sunlight per day, calculate the minimum solar panel surface area necessary to provide all of a typical home's electricity. (See Exercise 124 for the energy rate supplied by the sun.)

Short Answer

Expert verified
The minimum solar panel surface area necessary to provide all of a typical home's electricity is \(33.33 \, m^2\).

Step by step solution

01

Calculate the total energy required per day.

To find the energy consumption per day in Joules, we first need to convert the energy consumption from kilowatt-hours (kWh) to Joules. We'll use the following conversion factors: 1 kWh = 1 kW * 1 hour, and 1 kW = 1000 J/s. Total energy required per day (in Joules) = 40 kWh/day * 1 kW / 1 kWh * 3600 s/hour Total energy required per day = 144,000 kJ/day
02

Calculate the energy provided by sunlight per day.

Next, we need to find out how much energy the sun provides per day. To do this, we use the given energy rate of 1.00 kW/m² (1.00 kJ/s/m²), multiplied by the number of seconds of useful sunlight per day (8 hours * 3600 seconds/hour). Energy provided by sunlight per day (per m²) = 1.00 kJ/s/m² * 8 hours/day * 3600 s/hour Energy provided by sunlight per day (per m²) = 28,800 kJ/day/m²
03

Calculate the energy output of the solar panels per day.

Now, we need to calculate how much energy the solar panels can generate per day, considering their efficiency. The efficiency is given as 15%, which means that 15% of the sunlight energy is converted into electricity. Energy output from the solar panels per day (per m²) = energy provided by sunlight per day (per m²) * efficiency Energy output from the solar panels per day (per m²) = 28,800 kJ/day/m² * 0.15 Energy output from the solar panels per day (per m²) = 4,320 kJ/day/m²
04

Calculate the minimum solar panel surface area required.

Finally, we need to find the minimum surface area of solar panels required to provide all the energy needed for a typical home. To do this, we divide the total energy required per day by the energy output from the solar panels per day (per m²). Minimum solar panel surface area required = Total energy required per day / Energy output from the solar panels per day (per m²) Minimum solar panel surface area required = 144,000 kJ/day / 4,320 kJ/day/m² Minimum solar panel surface area required = 33.33 m² Thus, the minimum solar panel surface area necessary to provide all of a typical home's electricity is 33.33 m².

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Most popular questions from this chapter

One of the components of polluted air is NO. It is formed in the high- temperature environment of internal combustion engines by the following reaction: $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}(g) \quad \Delta H=180 \mathrm{kJ}$$ Why are high temperatures needed to convert \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) to NO?

Combustion reactions involve reacting a substance with oxygen. When compounds containing carbon and hydrogen are combusted, carbon dioxide and water are the products. Using the enthalpies of combustion for \(\mathrm{C}_{4} \mathrm{H}_{4}(-2341 \mathrm{kJ} / \mathrm{mol}), \mathrm{C}_{4} \mathrm{H}_{8}\) \((-2755 \mathrm{kJ} / \mathrm{mol}),\) and \(\mathrm{H}_{2}(-286 \mathrm{kJ} / \mathrm{mol}),\) calculate \(\Delta H\) for the reaction $$\mathrm{C}_{4} \mathrm{H}_{4}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{4} \mathrm{H}_{8}(g)$$

Given the following data $$\begin{aligned}\mathrm{Ca}(s)+2 \mathrm{C}(\text {graphite}) & \longrightarrow \mathrm{CaC}_{2}(s) & & \Delta H=-62.8 \mathrm{kJ} \\\ \mathrm{Ca}(s)+\frac{1}{2} \mathrm{O}_{2}(g) & \longrightarrow \mathrm{CaO}(s) & & \Delta H=-635.5 \mathrm{kJ} \\ \mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q) &\Delta H &=-653.1 \mathrm{kJ} \\\ \mathrm{C}_{2} \mathrm{H}_{2}(g)+\frac{5}{2} \mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \Delta H &=-1300 . \mathrm{kJ} \\ \mathrm{C}(\text {graphite})+\mathrm{O}_{2}(g) & \longrightarrow \mathrm{CO}_{2}(\mathrm{g}) & \Delta H &=-393.5 \mathrm{kJ} \end{aligned}$$ calculate \(\Delta H\) for the reaction $$\mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{C}_{2} \mathrm{H}_{2}(g)$$

A sample of nickel is heated to \(99.8^{\circ} \mathrm{C}\) and placed in a coffeecup calorimeter containing \(150.0 \mathrm{g}\) water at \(23.5^{\circ} \mathrm{C}\). After the metal cools, the final temperature of metal and water mixture is \(25.0^{\circ} \mathrm{C} .\) If the specific heat capacity of nickel is \(0.444 \mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g}\) what mass of nickel was originally heated? Assume no heat loss to the surroundings.

You have a 1.00 -mole sample of water at \(-30 .^{\circ} \mathrm{C}\) and you heat it until you have gaseous water at \(140 .^{\circ} \mathrm{C}\). Calculate \(q\) for the entire process. Use the following data. Specific heat capacity of ice \(=2.03 \mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g}\) Specific heat capacity of water \(=4.18 \mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g}\) Specific heat capacity of steam \(=2.02 \mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g}\) $$\begin{array}{ll}\mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) & \Delta H_{\text {fusion }}=6.02 \mathrm{kJ} / \mathrm{mol}\left(\mathrm{at} 0^{\circ} \mathrm{C}\right) \\\\\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g) & \Delta H_{\text {vaporization }}=40.7 \mathrm{kJ} / \mathrm{mol}\left(\text { at } 100 .^{\circ} \mathrm{C}\right)\end{array}$$
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