Question

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.

Answer 1

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)

Step-by-step 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

`\displaystyle 4186 \ J/kg \ ^\circ C((1000 \ g)/(1 \ kg)) = 4186000 \ J/g \ ^\circ C`

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

`\displaystyle 1.50741 \cdot 10^9 \ J((1 \ kJ)/(1000 \ J)) = 1.50741 \cdot 10^6 \ kJ`

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