Chemistry worksheet on calculating heat and specific heat, including formulas, a table of specific heat values, and practice problems.

Let’s solve each problem step by step.

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Problem 1:

Ethanol has a specific heat of 2.44 J/g°C.
Mass = 34.4 g
Temperature increases from 25°C to 78.8°C → so ΔT = 78.8 - 25 = 53.8°C

We use the formula:
q = m × c × ΔT

Plug in the numbers:
q = 34.4 g × 2.44 J/g°C × 53.8 °C

First, multiply 34.4 × 2.44:

34.4 × 2.44
= (34 × 2.44) + (0.4 × 2.44)
= 82.96 + 0.976 = 83.936

Now multiply that by 53.8:

83.936 × 53.8

Break it down:

83.936 × 50 = 4196.8
83.936 × 3.8 = let’s calculate:

83.936 × 3 = 251.808
83.936 × 0.8 = 67.1488
→ 251.808 + 67.1488 = 318.9568

Total q = 4196.8 + 318.9568 = 4515.7568 J

Round to reasonable sig figs — all inputs have 3 sig figs (34.4, 2.44, 53.8), so answer should be 4520 J or 4.52 × 10³ J

But since the problem doesn’t specify rounding, we’ll keep it as 4516 J if needed — but let’s check with calculator-style precision:

Actually, better to do direct multiplication:

34.4 × 2.44 = 83.936
83.936 × 53.8 = ?

Do 83.936 × 53.8:

= 83.936 × (50 + 3 + 0.8)
= 83.936×50 = 4196.8
83.936×3 = 251.808
83.936×0.8 = 67.1488
Sum: 4196.8 + 251.808 = 4448.608; +67.1488 = 4515.7568

So ≈ 4516 J

But let’s see what the worksheet expects — probably rounded to whole number or one decimal? Since mass and temp are given to 1 decimal, and specific heat to 2 decimals, maybe report as 4516 J or 4520 J.

Wait — actually, 34.4 has 3 sig figs, 2.44 has 3, 53.8 has 3 → product should have 3 sig figs.

4515.7568 → rounds to 4520 J (which is 4.52 × 10³)

But sometimes worksheets accept the full calculation. Let’s go with 4516 J for now — but I’ll double-check.

Alternatively, use exact:

34.4 * 2.44 = 83.936
83.936 * 53.8 = let's compute:

83.936 * 53.8
= (83936/1000) * (538/10) = (83936 * 538) / 10000

Too messy. Use calculator logic:

34.4 × 2.44 = 83.936
83.936 × 53.8 = 4515.7568 → 4516 J when rounded to nearest joule.

I think 4516 J is fine.

But wait — let me verify with another method:

ΔT = 78.8 - 25 = 53.8 ✔️
m = 34.4 ✔️
c = 2.44 ✔️

q = 34.4 * 2.44 * 53.8

Compute step-by-step:

Step 1: 34.4 * 2.44
34.4 * 2 = 68.8
34.4 * 0.4 = 13.76
34.4 * 0.04 = 1.376
Better: 2.44 = 2 + 0.4 + 0.04
So 34.4*2 = 68.8
34.4*0.4 = 13.76
34.4*0.04 = 1.376
Sum: 68.8 + 13.76 = 82.56; +1.376 = 83.936 ✔️

Step 2: 83.936 * 53.8

Write 53.8 as 50 + 3.8

83.936 * 50 = 4196.8
83.936 * 3.8:

83.936 * 3 = 251.808
83.936 * 0.8 = 67.1488
Sum: 251.808 + 67.1488 = 318.9568

Total: 4196.8 + 318.9568 = 4515.7568 → 4516 J

Yes.

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Problem 2:

Mass of Aluminum = 3 grams
Heated from 20°C to 662°C → ΔT = 662 - 20 = 642°C
Heat absorbed q = 1728 J

Find specific heat c.

Formula:
c = q / (m × ΔT)

Plug in:

c = 1728 J / (3 g × 642 °C)

First, denominator: 3 × 642 = 1926

So c = 1728 / 1926

Simplify fraction:

Divide numerator and denominator by 6:

1728 ÷ 6 = 288
1926 ÷ 6 = 321

Again divide by 3:

288 ÷ 3 = 96
321 ÷ 3 = 107

So 96/107 ≈ ?

Calculate decimal: 96 ÷ 107 ≈ 0.897...

Check original division: 1728 ÷ 1926

Do 1728 ÷ 1926:

≈ 0.8972

But let’s compute exactly:

1728 ÷ 1926 = ?

Use calculator steps:

1926 × 0.9 = 1733.4 → too big (since 1733.4 > 1728)

1926 × 0.897 = ?

1926 × 0.8 = 1540.8
1926 × 0.09 = 173.34
1926 × 0.007 = 13.482
Sum: 1540.8 + 173.34 = 1714.14; +13.482 = 1727.622 → very close to 1728

Difference: 1728 - 1727.622 = 0.378

So add 0.378 / 1926 ≈ 0.000196 → total ≈ 0.8972

So c ≈ 0.897 J/g°C

But look at the table — Aluminum is listed as 0.902 J/g°C — so our calculated value is close, which makes sense because this is likely based on real data.

But we must use the given numbers.

So c = 1728 / (3 × 642) = 1728 / 1926

Reduce fraction:

GCD of 1728 and 1926.

Factor 1728: 1728 = 2^6 × 3^3
1926: even → 963 × 2; 963 ÷ 3 = 321; 321 ÷ 3 = 107 → so 1926 = 2 × 3² × 107

Common factors: 2 × 3² = 18

So 1728 ÷ 18 = 96
1926 ÷ 18 = 107

So c = 96/107 J/g°C ≈ 0.897 J/g°C

To three significant figures? Mass is 3 g (1 sig fig?), but 3 could be exact. Temperature change 642 has 3 sig figs, heat 1728 has 4. Probably report as 0.897 J/g°C

But let’s see: 3 grams — if it’s exactly 3, then sig figs determined by 1728 and 642.

1728 has 4, 642 has 3 → so answer should have 3 sig figs.

0.897 has 3 sig figs → perfect.

So 0.897 J/g°C

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Problem 3:

Metal mass = 5 grams
Absorbs 240 J of heat
Temperature increases by 200°C → so ΔT = 200°C

Find specific heat c.

Formula:
c = q / (m × ΔT)

Plug in:

c = 240 J / (5 g × 200 °C) = 240 / 1000 = 0.24 J/g°C

Simple!

Check units: J / (g·°C) → correct.

Sig figs: 5 g (maybe 1 sig fig?), 240 J (2 or 3? If trailing zero not significant, maybe 2), 200°C (could be 1, 2, or 3 — ambiguous). But since 200 is written without decimal, often considered 1 sig fig — but in context, probably 3.

However, 5 × 200 = 1000 — which is exact if 5 and 200 are exact.

But 240 / 1000 = 0.24 — two sig figs.

If 5 is exact (like counted), and 200 is measured, then 200 has 3 sig figs? Unlikely — usually 200 without decimal is 1 sig fig.

But in school problems, they often treat such numbers as having implied precision.

Given that 240 has 2 or 3 sig figs — let’s assume 3 (if it’s 240. then 3).

But safest: 240 / (5 × 200) = 240 / 1000 = 0.24 → 0.24 J/g°C

And looking at the table, Gold is 0.129, Copper 0.385 — so 0.24 is between them, plausible.

So answer is 0.24 J/g°C

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Now, also there was an example to fill in:

Example:
50 g gold, c = 0.129 J/g°C, heated to 115°C, cools to 29.3°C → so ΔT = final - initial? Wait — it says “heated to 115°C; the gold cools until final temperature is 29.3°C”

So if it’s cooling, then ΔT = T_final - T_initial = 29.3 - 115 = -85.7°C

But heat released would be negative, but often they ask for magnitude.

The question says: “Calculate the Heat of the Metal” and gives q = m c ΔT

In the blank: q metal = _______ J

Since it’s cooling, heat is lost, so q should be negative.

But sometimes they want absolute value. Let’s see.

ΔT = 29.3 - 115 = -85.7°C

q = 50 g × 0.129 J/g°C × (-85.7 °C)

First, 50 × 0.129 = 6.45

Then 6.45 × (-85.7) = ?

6.45 × 85.7

6 × 85.7 = 514.2
0.45 × 85.7 = let’s compute: 0.4×85.7=34.28; 0.05×85.7=4.285; sum 38.565

Total: 514.2 + 38.565 = 552.765

So q = -552.765 J ≈ -553 J

But perhaps they want positive, meaning magnitude of heat released.

Looking at the worksheet, it says “the gold cools”, so heat is leaving the metal, so q is negative.

But in many introductory contexts, they might expect the absolute value.

However, the formula includes sign via ΔT.

I think we should include the sign.

But let’s check the instruction: “Calculate the Heat of the Metal” — ambiguous.

In physics, q is heat added to system, so if cooling, q < 0.

But perhaps for this level, they want the amount of heat transferred, so positive.

Looking at the example setup: they give q = m c ΔT, and ΔT is change, which can be negative.

But in the blank, it just says “q metal = _______ J”

I think safest is to compute with sign.

So q = 50 * 0.129 * (29.3 - 115) = 50 * 0.129 * (-85.7)

As above, ≈ -552.765 → round to -553 J

But let’s compute exactly:

50 × 0.129 = 6.45

6.45 × 85.7:

6.45 × 80 = 516
6.45 × 5.7 = 6.45×5 = 32.25; 6.45×0.7=4.515; sum 36.765
Total: 516 + 36.765 = 552.765

So -552.765 J → typically rounded to -553 J

But perhaps they expect positive, as "heat" sometimes means magnitude.

Given that in Problem 1, it asks "how much heat was absorbed", implying positive for absorption.

Here, it says "cools", so heat is released, so q is negative.

I'll go with -553 J

But let me see if the worksheet has any clue — no, it's blank.

Perhaps in the context, since it's "heat of the metal", and metal is losing heat, but still, q is defined as heat gained by system.

Standard convention: q > 0 if heat added to system.

So for the metal, as it cools, heat is removed, so q < 0.

So I'll put -553 J

But to match common practice in some textbooks, they might want 553 J with note that it's released.

Since the blank is just "q metal = _______ J", and no specification, I'll use the signed value.

Final decision: -553 J

But let's confirm calculation once more:

m = 50 g
c = 0.129 J/g°C
ΔT = T_final - T_initial = 29.3 - 115 = -85.7°C

q = 50 * 0.129 * (-85.7) = first 50*0.129=6.45

6.45 * 85.7:

85.7 * 6 = 514.2
85.7 * 0.45 = 85.7 * 0.4 = 34.28; 85.7 * 0.05 = 4.285; sum 38.565
Total 514.2 + 38.565 = 552.765

So -552.765 → rounds to -553 J (to nearest joule)

Yes.

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Now, compiling answers:

For the example: q metal = -553 J

Problem 1: 4516 J (or 4520? But we calculated 4515.7568, so 4516 is fine)

Earlier I thought about sig figs — 34.4 (3), 2.44 (3), 53.8 (3) → product should have 3 sig figs.

4516 has 4, so round to 4520 J? 4.52 × 10³ J

4515.7568 — to 3 sig figs: look at first three digits: 451, next digit is 5, so round up → 4520 J

Yes, 4520 J

Similarly, Problem 2: 0.897 J/g°C — 3 sig figs, good.

Problem 3: 0.24 J/g°C — 2 sig figs, since 5 and 200 may limit, but 240 has 2 or 3 — if 240 is 2 sig figs (trailing zero not significant), then 0.24 has 2.

But 5 g — if it's 5.0, it would be 2, but written as 5, possibly 1 sig fig. However, in such problems, often treated as exact.

To be safe, 0.24 is acceptable.

Some might say 0.240, but no.

I think 0.24 J/g°C is fine.

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Final Answers:

Example: q metal = -553 J

Problem 1: 4520 J

Problem 2: 0.897 J/g°C

Problem 3: 0.24 J/g°C

But let's write them clearly.

Also, for Problem 1, the question is "How much Heat was absorbed?" — so since temperature increased, heat absorbed, so positive, and we have 4520 J.

For Problem 2, specific heat is always positive, so 0.897 J/g°C.

Problem 3, same.

Example: since it cooled, heat was released, so q is negative.

But in some curricula, they might expect the magnitude. Given that the formula is given with ΔT, and ΔT is negative, I'll stick with negative.

Perhaps the worksheet expects positive for all, but logically, it should be negative.

Another way: sometimes "heat of the metal" means the heat content change, which is negative when cooling.

I think -553 J is correct.

But to match common homework expectations, let me see if there's a standard.

Upon second thought, in many high school worksheets, when they say "calculate the heat" for cooling, they might want the absolute value, especially if not specifying direction.

But the formula includes sign.

Looking back at the user's image description — it says "the gold cools until the final temperature is 29.3°C", and "calculate the Heat of the Metal".

In thermodynamics, q is heat transfer to the system, so for the metal, q < 0.

I'll keep it as -553 J.

But let's box the answers as per requirement.

Final Answer section should have all answers.

The user didn't specify which part, but since it's a worksheet, probably all blanks.

The example has a blank: q metal = _______ J

Then Problems 1,2,3.

So I need to provide:

For example: -553

Problem 1: 4520

Problem 2: 0.897

Problem 3: 0.24

With units as specified.

In Final Answer, I should list them.

Since the instruction is to provide final answer after reasoning, and no specific format, I'll write:

Final Answer:
Example: q metal = -553 J
Problem 1: 4520 J
Problem 2: 0.897 J/g°C
Problem 3: 0.24 J/g°C

But to be precise, let's ensure calculations are correct.

Recheck Problem 1 sig figs:

m = 34.4 g (3 sig figs)
c = 2.44 J/g°C (3 sig figs)
ΔT = 78.8 - 25 = 53.8°C (78.8 has 3, 25 has 2 — subtraction: 78.8 - 25.0 = 53.8, so 3 sig figs)

Product: 34.4 × 2.44 × 53.8

Each has 3 sig figs, so answer should have 3.

34.4 × 2.44 = 83.936 → but intermediate, keep extra digits.

83.936 × 53.8 = 4515.7568

Rounded to 3 sig figs: 4520 J (since 4515.7568, the third digit is 1, fourth is 5, so round up 451 to 452, so 4520)

Yes.

Problem 2: q = 1728 J (4 sig figs), m = 3 g (1 sig fig? But likely exact), ΔT = 642°C (3 sig figs)

If m=3 is exact, then limited by q and ΔT — q has 4, ΔT has 3, so c has 3 sig figs.

1728 / (3 * 642) = 1728 / 1926 = 0.897196... → 0.897 J/g°C

Problem 3: m=5 g (exact?), q=240 J (2 or 3 sig figs), ΔT=200°C (1,2, or 3)

If 200 has 3 sig figs (i.e., 200.), then 5*200=1000 exact, q=240 has 2 sig figs (if no decimal), so c=0.24 J/g°C with 2 sig figs.

If 240 has 3, then 0.240, but usually written as 0.24.

I think 0.24 is fine.

Example: m=50 g (2 sig figs? 50 could be 1 or 2 — if 50., then 2), c=0.129 (3 sig figs), ΔT=29.3-115= -85.7 (29.3 has 3, 115 has 3, difference 85.7 has 3)

So q = 50 * 0.129 * 85.7 — 50 has 2 sig figs, others have 3, so answer should have 2 sig figs.

50 * 0.129 = 6.45 → but 50 has 2 sig figs, so 6.5? No, multiplication: 50 has uncertainty, but typically in such cases, 50 is considered to have 2 sig figs.

50 × 0.129 = 6.45, which with 2 sig figs is 6.5? But 50 could be exact.

This is ambiguous.

50 g — if it's 50. g, then 2 sig figs, else 1.

In most problems, 50 is taken as 2 sig figs.

c=0.129 has 3, ΔT=85.7 has 3.

So limiting is 50 with 2 sig figs.

So q should have 2 sig figs.

|q| = 552.765 → to 2 sig figs: 550 J or 5.5 × 10² J

But 552.765 rounded to 2 sig figs is 550 J.

Since it's negative, -550 J.

But earlier I had -553, which is 3 sig figs.

Conflict.

Let's clarify sig fig rules for multiplication.

The number with the least number of significant figures determines the result.

Here, m=50 — if written as 50, it may have 1 or 2 sig figs. In scientific context, often assumed to have 2 if no decimal, but strictly, trailing zero without decimal is ambiguous.

In this worksheet, other masses are like 34.4, 3, 5 — so 50 is likely 2 sig figs.

c=0.129 has 3, ΔT=85.7 has 3.

So m has 2, so q has 2 sig figs.

50 * 0.129 = 6.45, but with 2 sig figs, it should be 6.5? No, the product is 50 * 0.129 * 85.7.

First, 50 * 85.7 = 4285, which with 2 sig figs is 4300? Better to calculate numerical value.

50 * 0.129 = 6.45 — but 50 has 2 sig figs, 0.129 has 3, so product has 2 sig figs: 6.5 (since 6.45 rounds to 6.5)

Then 6.5 * 85.7 = 557.05, which with 2 sig figs is 560? This is messy.

Standard way: calculate the product and round at the end.

50 * 0.129 * 85.7 = 552.765

Now, the factor with least sig figs is 50, which has 2 sig figs.

So round 552.765 to 2 sig figs: 550 J (since 5.5 × 10²)

552.765 — first two digits are 5 and 5, third is 2<5, so 550 J.

But 550 has 2 sig figs (the zeros are placeholders).

So |q| = 550 J, and since cooling, q = -550 J

Similarly, for Problem 1, we have 4520 J with 3 sig figs, which is consistent.

For Problem 2, m=3 g — if 3 has 1 sig fig, then c should have 1 sig fig.

3 * 642 = 1926, q=1728, c=1728/1926≈0.897, with 1 sig fig would be 0.9 J/g°C

But that seems too rough, and the table has 0.902, so probably m=3 is exact or has more precision.

In many problems, small integers like 3,5 are considered exact.

Similarly, 50 in example might be exact.

To avoid overcomplicating, and since the worksheet likely expects the calculated values without strict sig fig enforcement, I'll go with the precise calculations as done initially.

For consistency, I'll use:

Example: -553 J (as calculated)

Problem 1: 4520 J (rounded to 3 sig figs)

Problem 2: 0.897 J/g°C

Problem 3: 0.24 J/g°C

And for the example, perhaps they expect positive, but I'll keep negative.

Another idea: in the example, it says "heated to 115°C; the gold cools until final temperature is 29.3°C" — so the heating is separate, but the cooling is what we're calculating for the heat change during cooling.

The question is "Calculate the Heat of the Metal" — probably during the cooling process.

So q for the metal during cooling is negative.

I think it's correct.

So final answers:

Final Answer:
Example: q metal = -553 J
Problem 1: 4520 J
Problem 2: 0.897 J/g°C
Problem 3: 0.24 J/g°C

But to write it neatly.