Question

Ice has a specific heat of 2090 J/(kg C) and water has a specific heat of 4186 J/(kg C). Water has a latent heat of fusion of 3.3x10^5 J/K and a latent heat of vaporization of 2.26x10^6 J/K. A 10 kg block of ice begins at -60 degrees C and is slowly heated until the entire mass is boiled to steam. How much energy is required to bring the ice to steam with a temperature of 100 degrees C?

Answer 1

To bring the ice to steam with a temperature of 100 degrees C, you need a total of 8,740,000 J of energy. This includes energy to heat the ice from -60 degrees C to 0 degrees C, energy to melt the ice, and energy to heat the water from 0 degrees C to 100 degrees C.

Step-by-step explanation:

To calculate the energy required to bring the ice to steam with a temperature of 100 degrees C, we need to consider the different phases and their respective energy changes. First, we calculate the energy required to bring the ice from -60 degrees C to 0 degrees C, which can be done using the specific heat of ice. Q1 = m * c * ΔT = (10 kg) * (2090 J/(kg C)) * (0-(-60) C) = 1,254,000 J.

Next, we calculate the energy required to melt the ice to water at 0 degrees C, which is given by the latent heat of fusion. Q2 = m * Lf = (10 kg) * (3.3x10^5 J/K) = 3,300,000 J.

Finally, we calculate the energy required to heat the water from 0 degrees C to 100 degrees C, which can be done using the specific heat of water. Q3 = m * c * ΔT = (10 kg) * (4186 J/(kg C)) * (100-0) C = 4,186,000 J. The total energy required is the sum of Q1, Q2, and Q3: Q_total = Q1 + Q2 + Q3 = 1,254,000 J + 3,300,000 J + 4,186,000 J = 8,740,000 J.

Answer 2

3.1 × 10⁷ J

Step-by-step explanation:

The total heat required is the sum of the heats required in each stage.

1) Solid: from -60°C to 0°C.

Q₁ = c(s) × m × ΔT = (2090 J/kg.°C) × 10 kg × (0°C - (-60°C)) = 1.3 × 10⁶ J

where,

c(s): specific heat of the solid

m: mass

ΔT: change in the temperature

2) Solid to liquid at 0°C

Q₂ = Qf × m = (3.3 × 10⁵ J/kg) × 10 kg = 3.3 × 10⁶ J

where,

Qf: latent heat of fusion

3) Liquid: from 0°C to 100°C

Q₃ = c(l) × m × ΔT = (4186J/kg.°C) × 10 kg × (100°C - 0°C) = 4.2 × 10⁶ J

where,

c(l): specific heat of the liquid

4) Liquid to gas at 100 °C

Q₄ = Qv × m = (2.26 × 10⁶ J/kg) × 10 kg = 2.26 × 10⁷ J

where,

Qv: latent heat of vaporization

Total heat

Q₁ + Q₂ + Q₃ + Q₄

1.3 × 10⁶ J + 3.3 × 10⁶ J + 4.2 × 10⁶ J + 2.26 × 10⁷ J = 3.1 × 10⁷ J