If one gram of solid ice at -20 degrees C is heated to gaseous water at 125 degrees C how much energy is required?
Worked same way as the next problem you posted.
you
To calculate the energy required to heat the ice from -20 degrees Celsius to water vapor at 125 degrees Celsius, we need to consider the three stages of phase changes and the heating of the water.
1. Heating the ice from -20°C to 0°C:
The specific heat capacity of ice is approximately 2.09 J/g°C. To raise 1 gram of ice from -20°C to 0°C, we need to calculate the energy using the formula:
Energy = mass × specific heat capacity × temperature change
= 1 g × 2.09 J/g°C × (0°C - (-20°C))
= 1 g × 2.09 J/g°C × 20°C
= 41.8 J
So, it takes 41.8 J of energy to heat the ice from -20°C to 0°C.
2. Melting the ice at 0°C:
To change the state of the ice from solid to liquid, we need to consider the heat of fusion. The heat of fusion of ice is 334 J/g. Since we have 1 gram of ice, the energy required for melting is:
Energy = mass × heat of fusion
= 1 g × 334 J/g
= 334 J
Therefore, it takes 334 J of energy to melt the 1 gram of ice at 0°C.
3. Heating the liquid water to 100°C:
The specific heat capacity of water is around 4.18 J/g°C. To heat 1 gram of water from 0°C to 100°C, we can calculate the energy using the same formula:
Energy = mass × specific heat capacity × temperature change
= 1 g × 4.18 J/g°C × (100°C - 0°C)
= 1 g × 4.18 J/g°C × 100°C
= 418 J
Hence, it takes 418 J of energy to heat the liquid water from 0°C to 100°C.
4. Boiling the water at 100°C:
To convert 1 gram of water at 100°C to water vapor (steam) at the same temperature, we need to consider the heat of vaporization. The heat of vaporization of water is approximately 2260 J/g.
Energy = mass × heat of vaporization
= 1 g × 2260 J/g
= 2260 J
So, it takes 2260 J of energy to vaporize the 1 gram of water at 100°C.
Overall, to heat 1 gram of solid ice at -20°C to gaseous water at 125°C, the total energy required would be the sum of the energies calculated for each stage:
Total Energy = Energy for heating ice + Energy for melting ice + Energy for heating water + Energy for boiling water
= 41.8 J + 334 J + 418 J + 2260 J
= 3053.8 J
Therefore, it would require approximately 3053.8 Joules of energy to complete the heating process.