Activity 1.3 1 Solar Hydrogen System Answer Key: Exact Answer & Steps

Ever tried to crack the “Activity 1.3 1 Solar‑Hydrogen System” worksheet and felt like you were staring at a foreign language?
You’re not alone. Most students hit a wall the first time they see the answer key—because the key itself reads like a chemistry textbook mashed with an engineering manual. The short version is: you need a map that talks the way you think, not the way the textbook talks.

Below is that map. That's why i’ve walked through every step, flagged the common traps, and tossed in a few practical shortcuts that actually save you time. Grab a pen, keep the worksheet handy, and let’s demystify this solar‑hydrogen system together.

What Is the Solar‑Hydrogen System in Activity 1.3 1?

In plain English, the activity asks you to model a tiny power plant that uses sunlight to split water into hydrogen and oxygen, then burns the hydrogen to make electricity again. It’s a closed loop: solar panels → electrolysis → hydrogen storage → fuel cell → electricity.

Think of it as a miniature version of what scientists hope to scale up for clean‑energy grids. The worksheet breaks the system into three parts:

1. Solar array – converts photons into DC voltage.
2. Electrolyzer – uses that voltage to split H₂O into H₂ and O₂.
3. Fuel cell – recombines H₂ and O₂ to generate usable power.

The “answer key” you’re after is basically a set of numbers that tell you how much current, voltage, and gas each component should produce under the given conditions.

Why It Matters / Why People Care

Why bother with a classroom exercise about hydrogen? In real terms, because the real world is already testing these concepts. Day to day, companies are building solar‑hydrogen farms in deserts, and governments are funding pilot projects to replace diesel generators on remote islands. If you can nail the calculations now, you’ll understand the efficiency trade‑offs that engineers wrestle with daily.

And here’s the thing—most textbooks gloss over the “why.” They give you a formula, you plug in numbers, you get an answer, and that’s it. In practice, you have to juggle:

• Solar irradiance fluctuations (clouds, angle of the sun).
• Electrolyzer efficiency loss (heat, over‑potential).
• Hydrogen compression costs (energy to store the gas).

When you see those variables pop up in the answer key, you’ll finally see the connection between a classroom problem and a real‑world design decision.

How It Works (Step‑by‑Step)

Below is the full workflow the worksheet expects you to follow. I’ve kept the original numbering but added my own “why this matters” notes.

1. Calculate Solar Power Input

Given: Solar irradiance = 800 W/m², panel area = 2 m², panel efficiency = 15 %.

Step:
(P_{solar}= \text{irradiance} \times \text{area} \times \text{efficiency})

Plugging in:
(P_{solar}=800 \times 2 \times 0.15 = 240 W)

Why it matters: This is the raw energy you have to work with. If you over‑estimate, the downstream numbers (hydrogen produced, fuel‑cell output) will be way off.

2. Convert Solar Power to Electrical Power

Solar panels produce DC voltage. The worksheet assumes a panel voltage of 30 V.

Current from panels:
(I_{panel}= \frac{P_{solar}}{V_{panel}} = \frac{240}{30} = 8 A)

Note: Real panels rarely stay at a single voltage; they have a “maximum power point.” The answer key uses the idealized value, which is fine for a classroom model.

3. Electrolysis – How Much Hydrogen Is Made?

The electrolyzer’s efficiency is given as 70 % (i.Which means e. , 70 % of electrical energy becomes chemical energy). The theoretical energy needed to split water is 237 kJ per mole of H₂.

Step A – Electrical energy per second (watts) to chemical energy:
(E_{chem}=P_{solar} \times \text{efficiency}=240 \times 0.70 = 168 W)

Step B – Convert watts to moles per second:
(168 J/s \div 237,000 J/mol ≈ 7.1 × 10^{-4}, \text{mol/s})

Step C – Mass of H₂ per second:
(7.1 × 10^{-4}, \text{mol/s} \times 2.016 g/mol ≈ 1.43 × 10^{-3}, g/s)

That’s roughly 5 g per hour of hydrogen—enough to power a small LED array for a few minutes.

4. Hydrogen Storage – Pressure and Volume

The worksheet assumes the hydrogen is stored at 350 bar in a 5‑liter tank.

Ideal gas law (simplified):
(n = \frac{P V}{R T})

Using (P=350 bar) (≈35 MPa), (V=0.005 m³), (R=8.314 J/mol·K), and room temperature (T=298 K):

(n ≈ \frac{35 × 10^{6} × 0.005}{8.314 × 298} ≈ 71 mol)

That’s ≈ 143 g of H₂—far more than the electrolyzer can produce in an hour, so the tank never fills up in this scenario. Think about it: the answer key reflects this by showing a “storage safety factor” of 2. 5.

5. Fuel Cell – Turning Hydrogen Back Into Electricity

Fuel‑cell efficiency is listed as 60 %. Here's the thing — the cell voltage under load is about 0. 7 V per cell, and the stack has 50 cells.

Total voltage:
(V_{stack}=0.7 V × 50 = 35 V)

For more on this topic, read our article on wiley jones zombie house flipping or check out which waves have some electrical properties and some magnetic properties.

Current from stored H₂:
First, calculate the maximum moles of H₂ you can feed per second given the storage amount. The answer key simplifies this to a constant current of 5 A.

Power output:
(P_{fuel}=V_{stack} \times I = 35 × 5 = 175 W)

Adjusted for efficiency:
(P_{usable}=175 W × 0.60 ≈ 105 W)

That’s the number you’ll see in the “output power” box of the answer key. Less friction, more output.

6. Overall System Efficiency

Now pull everything together:

[ \text{Overall efficiency} = \frac{P_{usable}}{P_{solar}} = \frac{105}{240} ≈ 44 % ]

That’s the figure the key lists under “system performance.” It tells you that more than half the solar energy is lost as heat, over‑potential, and compression work—exactly the kind of insight you need for real‑world design.

Common Mistakes / What Most People Get Wrong

1. Skipping the efficiency factor – It’s tempting to use the raw solar wattage for later steps. Forgetting the 70 % electrolyzer efficiency inflates hydrogen production by almost 30 %.

2. Mixing units – The worksheet flips between joules, kilojoules, and watts. I’ve seen students write “237 kJ/mol” but then divide by “237 J/mol,” which drops the answer by a factor of 1,000.

3. Assuming 100 % conversion in the fuel cell – The key explicitly states 60 % efficiency, but many forget to apply it to the final power calculation, ending up with 175 W instead of the correct 105 W.

4. Over‑looking the storage safety factor – The answer key includes a multiplier (2.5) to ensure the tank never exceeds its pressure rating. Ignoring it can lead to a “tank over‑pressure” error in the worksheet.

5. Treating the solar panel voltage as constant – Real panels have an IV curve. The worksheet’s 30 V is a simplification; if you try to plug in the actual open‑circuit voltage you’ll get a mismatch in the current calculation.

Practical Tips / What Actually Works

• Create a quick reference table. Write down each given efficiency, voltage, and area on a sticky note. When you move from solar to electrolyzer, you can glance at it without hunting through the problem statement.

• Round only at the end. Early rounding (e.g., 0.70 → 0.7) can compound errors. Keep as many decimal places as your calculator allows until you hit the final answer.

• Check the units with a “unit‑audit” line. Write the units under each term before you cancel them. It forces you to spot mismatches before they become full‑blown mistakes.

• Use a spreadsheet for the gas‑law step. Plug the pressure, volume, and temperature into Excel or Google Sheets; the formula =P*V/(R*T) will handle the exponent gymnastics for you.

• Validate with a sanity check. If the final power output is higher than the solar input, you’ve made a mistake. In our case, 105 W is comfortably lower than 240 W, so the numbers feel realistic.

• Remember the “big picture.” The activity isn’t just about crunching numbers; it’s about understanding where losses happen. When you see the overall efficiency of ~44 %, ask yourself: Which component is the biggest energy hog? (Answer: the electrolyzer’s over‑potential.)

FAQ

Q1: Why does the electrolyzer use 70 % efficiency instead of a higher number?
A: Laboratory‑scale electrolyzers can reach 80‑90 %, but the worksheet assumes a commercial, off‑the‑shelf unit that loses energy as heat and through imperfect membranes. The 70 % figure reflects a realistic, cost‑effective device.

Q2: Can I use 25 °C instead of 298 K for the gas‑law calculation?
A: Yes, but you must convert to Kelvin first (25 °C + 273 = 298 K). Skipping the conversion will give a pressure that’s off by about 9 %.

Q3: What if my solar irradiance is 1000 W/m² instead of 800 W/m²?
A: Re‑run the first step: (P_{solar}=1000 × 2 × 0.15 = 300 W). Everything downstream scales linearly, so the final usable power becomes roughly 131 W, and overall efficiency stays near 44 %.

Q4: Is the 350 bar storage pressure realistic for a small lab setup?
A: For a classroom demonstration, it’s high but not impossible. Real‑world hydrogen stations often use 350–700 bar. If you can’t reach that pressure, just note the reduced storage capacity in your answer.

Q5: How do I know if my answer key is wrong?
A: Cross‑check the final power output against the solar input. If the answer key shows > 240 W, it’s a typo. Also, verify that each efficiency factor (70 %, 60 %) is applied exactly once.

That’s it. You now have the full answer key broken down into bite‑size pieces, a list of the traps that trip most students, and a handful of shortcuts that actually save you time. Next time you open the worksheet, you won’t just be filling in numbers—you’ll be seeing the whole solar‑hydrogen story unfold.

Good luck, and may your calculations be ever in your favor.