Question

What would be the heat capacity expressed in kJ o C - 1 of a water bath containing 4.00 dm 3 of water? The specific heat of water is 4.814 J g- 1 o C- 1.

Answer 1

To determine the heat capacity of a water bath containing 4.00 dm³ of water, we need to convert the given information into consistent units.

Given: Volume of water (V) = 4.00 dm³ Specific heat of water (c) = 4.814 J g⁻¹ °C⁻¹

To calculate the heat capacity, we need to convert the volume of water to grams and the specific heat to kJ.

1. Convert the volume of water to grams: Since the density of water is 1 g/cm³, we can assume 1 dm³ of water is equal to 1000 g: Mass of water (m) = Volume of water × Density of water m = 4.00 dm³ × 1000 g/dm³ = 4000 g

2. Convert the specific heat to kJ: Specific heat (c) = 4.814 J g⁻¹ °C⁻¹ Specific heat (c) = 4.814 × 10⁻³ kJ g⁻¹ °C⁻¹

3. Calculate the heat capacity: Heat capacity (C) = mass × specific heat C = 4000 g × 4.814 × 10⁻³ kJ g⁻¹ °C⁻¹

   C ≈ 19.256 kJ/°C

Therefore, the heat capacity of the water bath containing 4.00 dm³ of water is approximately 19.256 kJ/°C.