Worksheet answers for specific heat capacity calculations.

Let’s go through each problem step by step to make sure we get the right answers. We’ll use the formula:

Q = m × c × ΔT

Where:
- Q = heat energy (in joules, J)
- m = mass (in grams, g)
- c = specific heat capacity (in J/g°C)
- ΔT = change in temperature = T_final - T_initial (in °C)

We can rearrange this formula depending on what we’re solving for.

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Problem 1:

> A 15.75-g piece of iron absorbs 1086.75 joules of heat energy, and its temperature changes from 25°C to 175°C. Calculate the specific heat capacity of iron.

We are solving for c.

Rearranged formula:
c = Q / (m × ΔT)

ΔT = 175°C - 25°C = 150°C

Plug in values:
c = 1086.75 J / (15.75 g × 150°C)
First, multiply denominator: 15.75 × 150 = 2362.5

Now divide:
1086.75 ÷ 2362.5 = 0.46 J/g°C

✔ Correct as shown.

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Problem 2:

> How many joules of heat are needed to raise the temperature of 10.0 g of aluminum from 22°C to 55°C, if the specific heat of aluminum is 0.90 J/g°C?

We are solving for Q.

Formula:
Q = m × c × ΔT

ΔT = 55°C - 22°C = 33°C

Plug in:
Q = 10.0 g × 0.90 J/g°C × 33°C
First: 10.0 × 0.90 = 9.0
Then: 9.0 × 33 = 297 J

✔ Correct as shown.

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Problem 3:

> Calculate the specific heat capacity of a piece of wood if 1500.0 g of the wood absorbs 67,500 joules of heat, and its temperature changes from 32°C to 57°C.

Solving for c.

Formula:
c = Q / (m × ΔT)

ΔT = 57°C - 32°C = 25°C

Plug in:
c = 67500 J / (1500.0 g × 25°C)
Denominator: 1500 × 25 = 37,500

Divide:
67500 ÷ 37500 = 1.8 J/g°C

✔ Correct as shown.

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Problem 4:

> 100.0 g of 4.0°C water is heated until its temperature is 37°C. Calculate the amount of heat energy needed to cause this rise in temperature.

We need the specific heat of water — it’s 4.184 J/g°C (standard value).

Solving for Q.

Formula:
Q = m × c × ΔT

ΔT = 37°C - 4.0°C = 33.0°C

Plug in:
Q = 100.0 g × 4.184 J/g°C × 33.0°C

First: 100.0 × 4.184 = 418.4
Then: 418.4 × 33.0 = ?

Let’s compute that carefully:

418.4 × 30 = 12,552
418.4 × 3 = 1,255.2
Total = 12,552 + 1,255.2 = 13,807.2 J

But wait — the answer key says 14,000 J.

That suggests they rounded the specific heat or the final answer.

If we use c = 4.18 J/g°C (sometimes rounded):

Q = 100 × 4.18 × 33 = 418 × 33

418 × 30 = 12,540
418 × 3 = 1,254
Total = 13,794 → still not 14,000.

What if they used c = 4.2 J/g°C?

Q = 100 × 4.2 × 33 = 420 × 33 = 13,860 → closer but not 14,000.

Wait — maybe they did:

100 × 4.184 × 33 = let's calculate exactly:

4.184 × 33 = 138.072
Then × 100 = 13,807.2 J

But the worksheet says 14,000 J — which is rounded to 2 significant figures? Let’s check sig figs.

Given:
- 100.0 g → 4 sig figs
- 4.0°C → 2 sig figs
- 37°C → 2 sig figs
→ So ΔT = 33°C → 2 sig figs
→ Specific heat of water is usually taken as exact or 4 sig figs.

The limiting factor is ΔT with 2 sig figs? But 33 has two digits — actually, since both temps are given to nearest whole number, difference is 33, which is exact enough.

Actually, looking at the problem again — perhaps the worksheet expects us to use c = 4.184, but round the final answer to 2 sig figs? 13,807 rounds to 14,000 J if we consider it as 1.4 × 10⁴ — which is 2 sig figs.

Yes — because 37°C and 4.0°C — the 4.0 has 2 sig figs, 37 has 2, so ΔT = 33 has 2 sig figs? Actually, subtraction doesn’t follow sig fig rules the same way — but often in these problems, if inputs have 2–3 sig figs, answer is rounded accordingly.

In any case, the worksheet shows 14,000 J, which is acceptable as a rounded answer.

So we’ll accept 14,000 J as correct per the context.

✔ Matches worksheet.

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Problem 5:

> 25.0 g of mercury is heated from 25°C to 155°C, and absorbs 455 joules of heat in the process. Calculate the specific heat capacity of mercury.

Solving for c.

Formula:
c = Q / (m × ΔT)

ΔT = 155°C - 25°C = 130°C

Plug in:
c = 455 J / (25.0 g × 130°C)

Denominator: 25.0 × 130 = 3,250

Divide:
455 ÷ 3250 = ?

Let’s compute:

455 ÷ 3250 = 0.14

Because: 3250 × 0.14 = 3250 × 0.1 = 325; 3250 × 0.04 = 130; total 325 + 130 = 455 ✔

So c = 0.14 J/g°C

✔ Correct as shown.

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All answers match the worksheet. The calculations are accurate when following standard rounding practices expected in high school science.

Final Answer:
All provided answers in the worksheet are correct:
1. 0.46 J/g°C
2. 297 J
3. 1.8 J/g°C
4. 14000 J
5. 0.14 J/g°C