alyssahockett4
23.12.2019 •
Physics
Suppose 9.30 3 105 j of energy are transferred to 2.00 kg of ice at 0°c. (a) calculate the energy required to melt all the ice into liquid water. (b) how much energy remains to raise the temperature of the liquid water? (c) determine the final temperature of the liquid water in celsius.
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Ответ:
(a) 672000 J
(b) 258300 J
(c) 30.75 °C
Explanation:
(a) Energy required to melt the ice = Latent heat of fusion of ice
Latent heat of fusion of ice (Q₁) = lm Equation 1
Where l = specific latent heat of fusion of ice, m = mass of ice
Given: m = 2.00 kg
Constant : l = 336000 J/kg
Substituting these values into equation 1,
Q₁ = 336000 × 2
Q₁ = 672000 J.
Energy required to melt the ice = 672000 J
(b) The Energy remains to raise the temperature of the liquid water = Total Energy - Energy requires to melt the ice.
Given: Total Energy = 9.303 × 10⁵ J, = 930300 J
Energy remain to raise the temperature of the liquid water =
930300 - 672000
Total Energy = 258300 J
Energy remain to raise the temperature of the liquid water = 258300 J.
(c): Q = cmΔT equation 2
Where c = specific heat capacity of water, m = mass of water, ΔT = change in temperature, Q = Energy required to change the temperature of liquid water
Making ΔT the subject of formula in the equation above,
ΔT = Q/cm Equation 3
Given: m = 2.00 kg, Q = 258300 J
Constant: C = 4200 J/kg. K
Substituting these values into equation 3
ΔT = 258300/(4200×2)
ΔT = 258300/8400
ΔT = 30.75 K
But ΔT = T₂ - T₁
Where T₁ initial temperature, T₂ = final Temperature
T₂ = T₁ + ΔT,
Given: T₁ = 0 °C
Therefore,
T₂ = 30.75 + 0 = 30.75
T₂ = 30.75 °C
Final temperature of water = 30.75 °C
Ответ:
acute
Explanation: