A 80 g ice cube at 0°C is heated until 55 g has become water at 100°C and 25 g has become steam at 100°C. How much energy was added to accomplish the transformation?
To find out how much energy was added to accomplish the transformation, we need to calculate the energy for each phase change.
1. Energy to heat ice from 0°C to 0°C: This phase change does not require any energy, as the temperature remains constant.
2. Energy to melt the ice: To calculate the energy required to melt the ice, we can use the formula Q = m * ΔHf, where Q represents the heat energy, m is the mass of the ice, and ΔHf is the heat of fusion. The heat of fusion for water is 334 J/g. Therefore, the energy required to melt the ice is Q = 80 g * 334 J/g.
3. Energy to heat the water from 0°C to 100°C: We can use the formula Q = m * c * ΔT, where Q represents the heat energy, m is the mass of the water, c is the specific heat capacity of water (4.18 J/g°C), and ΔT is the change in temperature. In this case, ΔT = 100°C - 0°C, so the energy required is Q = 55 g * 4.18 J/g°C * 100°C.
4. Energy to vaporize the water: To calculate the energy required to vaporize the water, we can use the formula Q = m * ΔHv, where Q represents the heat energy, m is the mass of the water, and ΔHv is the heat of vaporization. The heat of vaporization for water is 2260 J/g. Therefore, the energy required to vaporize the water is Q = 25 g * 2260 J/g.
Finally, to determine the total energy added to accomplish the transformation, we sum up the energies from each phase change:
Total energy = Energy to melt ice + Energy to heat water + Energy to vaporize water
Total energy = (80 g * 334 J/g) + (55 g * 4.18 J/g°C * 100°C) + (25 g * 2260 J/g).
By plugging in the values and performing the calculations, you can find the total energy added to accomplish the transformation.