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.

(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

490 n

f=125n a=2.5m/s2