Worksheet on the First Law of Thermodynamics and Heat Transfer with five practice problems.

I'll solve each problem step by step, carefully checking my calculations.

Problem 1:
Given:
- Specific heat of water (c) = 4.186 J/g°C
- Mass (m) = 31.209 g
- Initial temperature = 20.15°C
- Final temperature = 43.82°C

Formula: Q = mcΔT

First, find ΔT:
ΔT = 43.82 - 20.15 = 23.67°C

Now calculate Q:
Q = 31.209 × 4.186 × 23.67
Q = 31.209 × 4.186 = 130.641
Q = 130.641 × 23.67 = 3,092.27 J

Answer 1: 3,092 J (or 3,092.27 J)

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Problem 2:
Given:
- Heat capacity (c) = 0.451 J/g°C
- Heat absorbed (Q) = 34592.3 J
- Initial temperature = 12.25°C
- Final temperature = 71.95°C

Find mass (m).

Formula: Q = mcΔT, so m = Q/(cΔT)

First, find ΔT:
ΔT = 71.95 - 12.25 = 59.70°C

Now calculate m:
m = 34592.3 / (0.451 × 59.70)
m = 34592.3 / 26.9247
m = 1,284.78 g

Answer 2: 1,285 g (or 1,284.78 g)

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Problem 3:
Given:
- Initial temperature = 143.72°C
- Final temperature = 37.81°C
- Mass (m) = 293.3 g
- Heat transferred (Q) = 25208.3 J

Find specific heat (c).

Note: The sample cools down, so it loses energy. We use the absolute value.

Formula: Q = mcΔT, so c = Q/(mΔT)

First, find ΔT:
ΔT = 143.72 - 37.81 = 105.91°C

Now calculate c:
c = 25208.3 / (293.3 × 105.91)
c = 25208.3 / 31,063.403
c = 0.8115 J/g°C

Answer 3: 0.812 J/g°C (or 0.8115 J/g°C)

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Problem 4:
Given:
- Mass (m) = 0.9030 g
- Heat released (Q) = 448.24 J (released means negative in formula)
- Surroundings temperature = 9.49°C (this is the final temperature)
- Specific heat (c) = 0.91 J/g°C

Find initial temperature.

Formula: Q = mcΔT, where ΔT = T_final - T_initial

Since heat is released to surroundings, Q = -448.24 J
-448.24 = 0.9030 × 0.91 × (9.49 - T_initial)

First calculate mc:
mc = 0.9030 × 0.91 = 0.82173

So: -448.24 = 0.82173 × (9.49 - T_initial)

Divide both sides by 0.82173:
-448.24 / 0.82173 = 9.49 - T_initial
-545.48 = 9.49 - T_initial

Solve for T_initial:
T_initial = 9.49 + 545.48
T_initial = 554.97°C

Answer 4: 555°C (or 554.97°C)

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Problem 5:
Given:
- Mass (m) = 22.883 g
- Specific heat (c) = 1.05 J/g°C
- Initial temperature = 23.1°C
- Final temperature = 10.03°C

Find heat transferred (Q).

Formula: Q = mcΔT

First, find ΔT:
ΔT = 10.03 - 23.1 = -13.07°C

Now calculate Q:
Q = 22.883 × 1.05 × (-13.07)
Q = 22.883 × 1.05 = 24.027
Q = 24.027 × (-13.07) = -314.03 J

The negative sign indicates heat was released/lost.

Answer 5: -314 J (or -314.03 J), meaning 314 J of heat was released

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Final Answer:
1) 3,092 J
2) 1,285 g
3) 0.812 J/g°C
4) 555°C
5) -314 J (heat released)