Answer:
801.1 kJ
Explanation:
The ice increases in temperature from -20 °C to 0 °C and then melts at 0 °C.
The heat required to raise the ice to 0 °C is Q₁ = mc₁Δθ₁ where m = mass of ice = 1 kg, c₁ = specific heat capacity of ice = 2108 J/kg°C and Δθ₁ = temperature change. Q₁ = 1 kg × 2108 J/kg°C × (0 - (-20))°C = 2108 J/kg°C × 20 °C = 4216 J
The latent heat required to melt the ice is Q₂ = mL₁ where L₁ = specific latent heat of fusion of ice = 336000 J/kg. Q₁ = 1 kg × 336000 J/kg = 336000 J
The heat required to raise the water to 100 °C is Q₃ = mc₂Δθ₂ where m = mass of ice = 1 kg, c₂ = specific heat capacity of water = 4187 J/kg°C and Δθ₂ = temperature change. Q₃ = 1 kg × 4187 J/kg°C × (100 - 0)°C = 4187 J/kg°C × 100 °C = 418700 J
The latent heat required to convert the water to steam is Q₄ = mL₂ where L = specific latent heat of vapourisation of water = 2260 J/kg. Q₄ = 1 kg × 2260 J/kg = 2260 J
The heat required to raise the steam to 120 °C is Q₅ = mc₃Δθ₃ where m = mass of ice = 1 kg, c₃ = specific heat capacity of steam = 1996 J/kg°C and Δθ₃ = temperature change. Q₃ = 1 kg × 1996 J/kg°C × (120 - 100)°C = 1996 J/kg°C × 20 °C = 39920 J
The total amount of heat Q = Q₁ + Q₂ + Q₃ + Q₄ + Q₅ = 4216 J + 336000 J
+ 418700 J + 2260 J + 39920 J = 801096 J ≅ 801.1 kJ
I Believe the answer is A but D looks good too.
Answer:
A rocket with more mass will speed up more slowly, just as in the horizontal example, but there is another effect. The force of gravity is now acting in the opposite direction to the thrust, so the resultant force pushing the rocket upwards is also less.
Explanation:
Rosa is conducting the scientific practice of observation.
