Final answer:
To convert 10.0 g of ice at -20.0°C to steam at 110.0°C, a total of 8350 J of energy is required. This includes heating the ice from -20.0°C to its melting point, melting the ice, and then heating the water from the melting point to the final temperature.
Step-by-step explanation:
To calculate the energy required for this process, we need to consider the different steps involved in heating the ice and then converting it into steam. First, we need to heat the ice from -20.0°C to the melting point of the ice, which is 0.0°C. The amount of energy required for this step is given by the formula Q = m * C * AT, where Q is the energy, m is the mass of the ice, C is the specific heat capacity of ice, and AT is the change in temperature. In this case, m = 10.0 g, C = 2.06 J/g°C, and AT = 20.0°C. Plugging these values into the formula gives us Q = 10.0 g * 2.06 J/g°C * 20.0°C = 412 J.
Next, we need to calculate the energy required to melt the ice. The formula for this is Q = m * L, where Q is the energy, m is the mass of the ice, and L is the latent heat of fusion of ice. The latent heat of fusion of ice is 334 J/g. Plugging in the values, we get Q = 10.0 g * 334 J/g = 3340 J.
Finally, we need to calculate the energy required to heat the water from the melting point to the final temperature of 110.0°C. We use the same formula as before, Q = m * C * AT, with m = 10.0 g, C = 4.18 J/g°C (specific heat capacity of water), and AT = 110.0 - 0.0 = 110.0°C. Plugging in the values, we get Q = 10.0 g * 4.18 J/g°C * 110.0°C = 4598 J. Adding up the energy required for each step gives us a total energy of 412 J + 3340 J + 4598 J = 8350 J.