How many calories are absorbed when 56.0 g of liquid water at 100C is vaporized to steam at 100C?
q = mass x delta Hvap.
To calculate the number of calories absorbed when water is vaporized, you need to consider the heat energy required for phase change. The specific heat for liquid water and the heat of vaporization are involved in this calculation.
Here's how you can calculate it step by step:
1. Determine the heat energy required to raise the temperature of 56.0 g of liquid water from 0°C to 100°C. The specific heat capacity of liquid water is 1 calorie/(g°C). The formula to calculate heat energy is Q = m * C * ΔT, where Q is the heat energy, m is the mass of the substance, C is the specific heat capacity, and ΔT is the change in temperature. In this case, ΔT = 100°C - 0°C = 100°C. Thus, Q1 = 56.0 g * 1 calorie/(g°C) * 100°C.
2. Determine the heat energy required for the phase change from liquid water at 100°C to steam at 100°C. The heat of vaporization for water is 540 calories/g. The formula to calculate heat energy during a phase change is Q = m * H, where Q is the heat energy, m is the mass of the substance undergoing the phase change, and H is the heat of vaporization. In this case, m = 56.0 g, and H = 540 calories/g. Thus, Q2 = 56.0 g * 540 calories/g.
3. Calculate the total heat energy required by adding the two previous results: Q_total = Q1 + Q2.
By performing these calculations, you should be able to determine the number of calories absorbed when 56.0 g of liquid water at 100°C vaporizes to steam at 100°C.