How much heat is required to change 10 grams of ice at 0 deg. C to steam at 100 deg C?
answered above.
To find the amount of heat required to change the state of a substance, you need to consider the heat energy required for each step of the phase change. Here are the steps involved in changing 10 grams of ice at 0°C to steam at 100°C:
1. Heating the ice from -10°C to 0°C (Phase Change: Solid to Solid)
- The specific heat capacity of ice is 2.09 Joules per gram per degree Celsius (J/g°C).
- The temperature change is 0°C - (-10°C) = 10°C.
- The heat required can be calculated using the formula: Q = m * c * ΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the temperature change.
- Q = 10 g * 2.09 J/g°C * 10°C = 209 Joules.
2. Melting the ice at 0°C to water (Phase Change: Solid to Liquid)
- The heat of fusion for ice is 334 Joules per gram (J/g).
- The heat required to melt the ice can be calculated using the formula: Q = m * Hf, where Hf is the heat of fusion.
- Q = 10 g * 334 J/g = 3340 Joules.
3. Heating the water from 0°C to 100°C (Phase Change: Liquid to Liquid)
- The specific heat capacity of water is 4.18 Joules per gram per degree Celsius (J/g°C).
- The temperature change is 100°C - 0°C = 100°C.
- The heat required can be calculated using the formula: Q = m * c * ΔT.
- Q = 10 g * 4.18 J/g°C * 100°C = 4180 Joules.
4. Vaporizing the water at 100°C to steam (Phase Change: Liquid to Gas)
- The heat of vaporization for water is 2260 Joules per gram (J/g).
- The heat required to vaporize the water can be calculated using the formula: Q = m * Hv, where Hv is the heat of vaporization.
- Q = 10 g * 2260 J/g = 22600 Joules.
To find the total heat required to change 10 grams of ice at 0°C to steam at 100°C, you need to add the heat values from each step:
Total heat = 209 J + 3340 J + 4180 J + 22600 J = 30,329 Joules.
Therefore, it would require 30,329 Joules of heat to change 10 grams of ice at 0°C to steam at 100°C.