contestada

q=mcAt

Calculate the amount of heat gained by 114.32 grams of water at 14.85C raised to 18.0C. The heat capacity of water is 4186J/kgC.

Respuesta :

Space

Answer:

1.51 × 10⁶ kJ

General Formulas and Concepts:

Stoichiometry

  • Using Dimensional Analysis

Thermochemistry

Specific Heat Formula: q = mcΔT

  • q is heat (in J)
  • m is mass (in g)
  • c is specific heat formula (in J/g °C)
  • ΔT is change in temperature (in °C)

Explanation:

Step 1: Define

[Given] m = 114.32 g

[Given] c = 4186 J/kg °C

[Given] ΔT = 18.0 °C - 14.85 °C = 3.15 °C

[Solve] q

Step 2: Convert Pt. 1

1 kg = 1000 g

[tex]\displaystyle 4186 \ J/kg \ ^\circ C(\frac{1000 \ g}{1 \ kg}) = 4186000 \ J/g \ ^\circ C[/tex]

Step 3: Solve

  1. Substitute in variables [Specific Heat Formula]:                                            q = (114.32 g)(418600 J/g °C)(3.15 °C)
  2. Multiply:                                                                                                             q = (4.78544 × 10⁸ J/°C)(3.15 °C)
  3. Multiply:                                                                                                             q = 1.50741 × 10⁹ J

Step 4: Convert Pt. 2

1000 J = 1 kJ

[tex]\displaystyle 1.50741 \cdot 10^9 \ J(\frac{1 \ kJ}{1000 \ J}) = 1.50741 \cdot 10^6 \ kJ[/tex]

Step 5: Check

Follow sig fig rules and round. We are given 3 sig figs as our lowest.

1.50741 × 10⁶ kJ ≈ 1.51 × 10⁶ kJ

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