Chemistry worksheet on specific heat calculations with handwritten solutions and a reference chart of specific heat capacities.

Let’s solve Problem 2 step by step.

We are told:
- A 32.07 g sample of vanadium is heated to 75.00 °C (initial temp of metal).
- It’s dropped into a calorimeter with water at 22.50 °C (initial temp of water).
- Final temperature of everything = 26.30 °C.
- We need to find the mass of water in the calorimeter.

This is a heat transfer problem:
The hot vanadium loses heat → that heat is gained by the cooler water.
So:
Heat lost by vanadium = Heat gained by water

We use the formula:
Q = s × m × ΔT
where:
- Q = heat energy (in Joules)
- s = specific heat capacity (J/g·°C)
- m = mass (g)
- ΔT = change in temperature = T_final - T_initial

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Step 1: Find heat lost by vanadium (system)

From the chart, specific heat of vanadium (V) = 0.4886 J/g·°C

Mass of vanadium = 32.07 g
Initial temp of vanadium = 75.00 °C
Final temp = 26.30 °C
ΔT_vanadium = 26.30 - 75.00 = -48.70 °C (negative because it’s losing heat)

Q_vanadium = s × m × ΔT
= 0.4886 J/g·°C × 32.07 g × (-48.70 °C)

Let’s calculate that:

First, multiply 0.4886 × 32.07:

0.4886 × 32.07 ≈ let’s compute:

0.4886 × 30 = 14.658
0.4886 × 2.07 ≈ 1.0114
Total ≈ 14.658 + 1.0114 = 15.6694

Now multiply by -48.70:

15.6694 × 48.70 ≈ ?

Let’s do 15.6694 × 50 = 783.47
Minus 15.6694 × 1.3 = ~20.37
So 783.47 - 20.37 ≈ 763.1

But since it’s negative: Q_vanadium ≈ -763.1 J

Wait — let me recalculate more accurately using calculator-style steps:

Actually, better to do exact multiplication:

0.4886 × 32.07 = ?

Compute:

32.07 × 0.4886

Break it down:

32.07 × 0.4 = 12.828
32.07 × 0.08 = 2.5656
32.07 × 0.008 = 0.25656
32.07 × 0.0006 = 0.019242

Add them:

12.828
+2.5656 = 15.3936
+0.25656 = 15.65016
+0.019242 = 15.669402

Now multiply by 48.70:

15.669402 × 48.70

Do 15.669402 × 40 = 626.77608
15.669402 × 8 = 125.355216
15.669402 × 0.7 = 10.9685814

Add:

626.77608 + 125.355216 = 752.131296
+10.9685814 = 763.0998774

So Q_vanadium = -763.10 J (approximately)

But wait — in the student’s work, they used -760.0 J. Let’s check if we should round differently or if there was rounding earlier.

Actually, looking back at the student’s calculation:

They wrote:
Q = 0.4866 · 32.07 · 48.70 = 759.98 J → rounded to 760.0 J

Wait — they used 0.4866? But the table says vanadium is 0.4886!

That’s a mistake! The specific heat for vanadium is 0.4886, not 0.4866.

So their value is slightly off. But maybe it’s a typo in handwriting? In the table it clearly says 0.4886.

In their “SYSTEM” section, they wrote s = 0.4886, but then in the bottom calculation they wrote 0.4866 — probably a slip.

To be accurate, we’ll use 0.4886.

So Q_vanadium = 0.4886 × 32.07 × (-48.70) = -763.10 J

Therefore, heat gained by water = +763.10 J

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Step 2: Use this to find mass of water

For water:
s_water = 4.184 J/g·°C (from table)
ΔT_water = T_final - T_initial = 26.30 - 22.50 = +3.80 °C

Q_water = s × m × ΔT
763.10 = 4.184 × m × 3.80

Solve for m:

m = 763.10 / (4.184 × 3.80)

First, compute denominator:
4.184 × 3.80

4 × 3.80 = 15.2
0.184 × 3.80 = let’s compute:
0.18 × 3.80 = 0.684
0.004 × 3.80 = 0.0152
Total = 0.6992

So total denominator = 15.2 + 0.6992 = 15.8992

Now m = 763.10 / 15.8992 ≈ ?

Divide:

763.10 ÷ 15.8992

Estimate: 15.8992 × 48 = ?

15.8992 × 50 = 794.96
Minus 15.8992 × 2 = 31.7984 → 794.96 - 31.7984 = 763.1616

Wow — that’s almost exactly 763.10!

So 15.8992 × 48 = 763.1616 ≈ 763.10

So m ≈ 48.00 g

Wait — let’s check:

If m = 48.00 g, then Q = 4.184 × 48.00 × 3.80

Compute:

4.184 × 48 = ?

4 × 48 = 192
0.184 × 48 = 8.832
Total = 200.832

Then 200.832 × 3.80

200 × 3.80 = 760
0.832 × 3.80 ≈ 3.1616
Total = 763.1616 J — which matches our Q_vanadium magnitude.

Perfect.

So mass of water = 48.00 g

But in the student’s answer, they got 47.80 g — likely because they used 0.4866 instead of 0.4886 for vanadium’s specific heat.

Since the correct specific heat is 0.4886, we must use that.

Thus, final answer is 48.00 g.

But let’s confirm once more with precise numbers:

Q_lost_by_V = 0.4886 * 32.07 * (26.30 - 75.00)
= 0.4886 * 32.07 * (-48.70)
= - (0.4886 * 32.07 * 48.70)

Calculate 0.4886 * 32.07:

As before: 15.669402

Then 15.669402 * 48.70 = 763.0998774 → so |Q| = 763.10 J

Now for water:

m = Q / (s * ΔT) = 763.10 / (4.184 * 3.80)

4.184 * 3.80 = 15.8992

763.10 / 15.8992 = ?

Let me divide:

763.10 ÷ 15.8992

Do long division or use approximation:

15.8992 × 48 = 763.1616 as above

Difference: 763.1616 - 763.10 = 0.0616 too high

So adjust: subtract a tiny bit from 48.

Each 0.01 g less would reduce Q by 4.184 * 0.01 * 3.80 ≈ 0.159 J

We have excess 0.0616 J, so reduce mass by 0.0616 / 0.159 ≈ 0.387 g? Wait no — that can’t be right.

Wait — actually, since 15.8992 × 48 = 763.1616, and we want 763.10, difference is -0.0616

So m = 48 - (0.0616 / 15.8992) ≈ 48 - 0.00387 ≈ 47.996 g

Which rounds to 48.00 g

Yes.

Alternatively, use calculator precision:

763.10 / 15.8992 = 47.996... ≈ 48.00 g

So the correct mass is 48.00 grams

The student’s answer of 47.80 g came from using 0.4866 instead of 0.4886 — a small error in copying the specific heat value.

Final Answer:
48.00 g