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AveGali [126]
3 years ago
10

Water (3030 g ) is heated until it just begins to boil. If the water absorbs 5.49×105 J of heat in the process, what was the ini

tial temperature of the water?
Chemistry
1 answer:
Romashka-Z-Leto [24]3 years ago
7 0

The initial temperature of the water was 56.7 °C.

<em>Step 1</em>. Calculate the <em>temperature chang</em>e.

The formula for the heat (<em>q</em>) absorbed is

<em>q = mC</em>Δ<em>T </em>

We can solve this equation to get

Δ<em>T = q</em>/(<em>mC</em>)

<em>q</em> = 5.49 × 10⁵ J; <em>m</em> = 3030 g; <em>C</em> = 4.184 J·°C⁻¹g⁻¹

∴ Δ<em>T</em> = 5.49 × 10⁵ J/(3030 g × 4.184 J·°C⁻¹g⁻¹) = 43.30 °C

<em>Step </em>2. Calculate the <em>initial temperature </em>

Δ<em>T</em> = <em>T</em>_f – <em>T</em>_i = 100°C – <em>T</em>_i = 43.30 °C

<em>T</em>_i = 100 °C – 43.30 °C = 56.7 °C

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<h3>Answer:</h3>

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<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

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<u>Step 1: Define</u>

[Given] 3.2 × 10²⁴ atoms Si

[Solve] grams Si

<u>Step 2: Identify Conversions</u>

Avogadro's Number

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<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                       \displaystyle 3.2 \cdot 10^{24} \ atoms \ Si(\frac{1 \ mol \ Si}{6.022 \cdot 10^{23} \ atoms \ Si})(\frac{28.09 \ g \ Si}{1 \ mol \ Si})
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<u>Step 4: Check</u>

<em>Follow sig fig rules and round. Instructed to round to 2 sig figs.</em>

149.266 g Si ≈ 150 g Si

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