How much heat is required to warm 2.50 L of water from 25.0 °C to 100.0 °C? (Assume a density of 1.0 g/mL for the water.)

Verified step by step guidance

1. Convert the volume of water from liters to milliliters: 2.50 L = 2500 mL.
2. Use the density of water to find the mass: mass = volume \times density = 2500 \text{ mL} \times 1.0 \text{ g/mL}.
3. Calculate the temperature change (\Delta T): \Delta T = 100.0 \degree C - 25.0 \degree C.
4. Use the specific heat capacity of water (c = 4.18 \text{ J/g°C}) to set up the heat equation: q = m \times c \times \Delta T.
5. Substitute the values for mass, specific heat capacity, and temperature change into the equation to find the heat (q) required.

Key Concepts

Specific Heat Capacity

Specific heat capacity is the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius. For water, this value is approximately 4.18 J/g°C. Understanding this concept is crucial for calculating the heat required to change the temperature of water, as it directly relates the mass of the water and the temperature change to the total heat absorbed or released.

Mass of Water

To calculate the heat required, we first need to determine the mass of the water. Given the volume of water (2.50 L) and the density (1.0 g/mL), we can convert the volume to mass using the formula: mass = volume × density. This step is essential because the specific heat capacity is defined per gram, so knowing the mass allows us to apply the specific heat formula accurately.

Heat Transfer Equation

The heat transfer equation, often expressed as q = m × c × ΔT, relates the heat (q) absorbed or released to the mass (m) of the substance, its specific heat capacity (c), and the change in temperature (ΔT). In this scenario, ΔT is the difference between the final and initial temperatures of the water. This equation is fundamental for solving the problem, as it provides a direct method to calculate the total heat required for the temperature change.