Calculating the Energy Required to Heat Water from -14°C to 123°C at 1 ATM

To calculate the total energy required to heat 65.0 grams of water (H2O) from -14.0 °C to 123.0 °C at 1 ATM, we need to consider several steps in the process:

Heating the ice from -14.0 °C to 0 °C Melting the ice at 0 °C to liquid water Heating the liquid water from 0 °C to 100 °C Vaporizing the water at 100 °C to steam Heating the steam from 100 °C to 123.0 °C

These steps are collectively known as sensible heat, phase change, and latent heat. To perform this calculation accurately, we will use specific heat constants and phase change energies. These constants are essential for ensuring the calculation is precise.

Constants Needed

Specific heat of ice (cice): 2.09 J/g°C Specific heat of water (cwater): 4.18 J/g°C Specific heat of steam (csteam): 2.01 J/g°C Heat of fusion of water (ΔHfus): 334 J/g Heat of vaporization of water (ΔHvap): 2260 J/g

Step-by-Step Calculation

Step 1: Heating ice from -14.0 °C to 0 °C

q1 m cice ΔT 65.0 g 2.09 J/g°C (0 - -14.0 °C) q1 65.0 2.09 14.0 1914.1 J

Step 2: Melting ice at 0 °C to liquid water

q2 m ΔHfus 65.0 g 334 J/g q2 21710 J

Step 3: Heating water from 0 °C to 100 °C

q3 m cwater ΔT 65.0 g 4.18 J/g°C (100 - 0 °C) q3 65.0 4.18 100 27170 J

Step 4: Vaporizing water at 100 °C to steam

q4 m ΔHvap 65.0 g 2260 J/g q4 147900 J

Step 5: Heating steam from 100 °C to 123.0 °C

q5 m csteam ΔT 65.0 g 2.01 J/g°C (123.0 - 100 °C) q5 65.0 2.01 23.0 3026.95 J

Total Energy Required

Now, summing all the energy values calculated:

qtotal q1 q2 q3 q4 q5
qtotal 1914.1 J 21710 J 27170 J 147900 J 3026.95 J
qtotal 201721.05 J

Final Answer: The total energy required to heat 65.0 grams of water from -14.0 °C to 123.0 °C at 1 ATM is approximately 201721 J or 201.7 kJ.