Answer:
initial: 1654.6 J, final: 0 J, change: -1654.6 J
Explanation:
The length of the slide is
L = 8.80 ft = 2.68 m
So the height of the child when he is at the top of the slide is (with respect to the ground)
The potential energy of the child at the top is given by:
where
m = 63.0 kg is the mass of the child
g = 9.8 m/s^2 is the acceleration due to gravity
h = 1.13 m
Substituting,
At the bottom instead, the height is zero:
h = 0
So the potential energy is also zero: U = 0 J.
This means that the change in potential energy as the child slides down is
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
Answer:
maybe phone cause matter is a solid liquid or gad
