Question

How much heat energy is required to change a 0.3 kg ice cube from a solid at -20°C to steam at 120°C?

Answer 1

To calculate the total heat energy required to change the state of a substance from solid to steam, you need to consider the specific heat capacity and latent heat of the substance at different phases. Here's the breakdown of the heat energy required for the given scenario:

1. Heating the ice from -20°C to 0°C: The specific heat capacity of ice is approximately 2,093 J/kg°C. To raise the temperature of the ice from -20°C to its melting point at 0°C, we need to calculate the heat energy for this temperature range: Q1 = mass × specific heat capacity × temperature change Q1 = 0.3 kg × 2093 J/kg°C × (0°C - (-20°C)) Q1 = 0.3 kg × 2093 J/kg°C × 20°C Q1 = 12,558 J

2. Melting the ice at 0°C to water at 0°C: To change the state of the ice at 0°C to water at 0°C, we need to consider the latent heat of fusion. The latent heat of fusion for water is approximately 334,000 J/kg. Q2 = mass × latent heat of fusion Q2 = 0.3 kg × 334,000 J/kg Q2 = 100,200 J

3. Heating the water from 0°C to 100°C: The specific heat capacity of water is approximately 4,186 J/kg°C. To raise the temperature of the water from 0°C to its boiling point at 100°C, we calculate: Q3 = mass × specific heat capacity × temperature change Q3 = 0.3 kg × 4186 J/kg°C × (100°C - 0°C) Q3 = 0.3 kg × 4186 J/kg°C × 100°C Q3 = 125,580 J

4. Boiling the water at 100°C to steam at 100°C: To change the state of the water at 100°C to steam at 100°C, we need to consider the latent heat of vaporization. The latent heat of vaporization for water is approximately 2,260,000 J/kg. Q4 = mass × latent heat of vaporization Q4 = 0.3 kg × 2,260,000 J/kg Q4 = 678,000 J

5. Heating the steam from 100°C to 120°C: The specific heat capacity of steam is approximately 2,030 J/kg°C. To raise the temperature of the steam from 100°C to 120°C, we calculate: Q5 = mass × specific heat capacity × temperature change Q5 = 0.3 kg × 2030 J/kg°C × (120°C - 100°C) Q5 = 0.3 kg × 2030 J/kg°C × 20°C Q5 = 12,180 J

Total heat energy: Total Q = Q1 + Q2 + Q3 + Q4 + Q5 Total Q = 12,558 J + 100,200 J + 125,580 J + 678,000 J + 12,180 J Total Q ≈ 928,518 J

Therefore, approximately 928,518 Joules of heat energy are required to change a 0.3 kg ice cube from a solid at -20°C to steam at 120°C.