Question

An open container holds ice of mass 0.560 kg at a temperature of -19.6 ∘C . The mass of the container can be ignored. Heat is supplied to the container at the constant rate of 760 J/minute . The specific heat of ice to is 2100 J/kg⋅K and the heat of fusion for ice is 334×103J/kg.

a. How much time tmelts passes before the ice starts to melt?

b. From the time when the heating begins, how much time trise does it take before the temperature begins to rise above 0∘C?

Answer 1

Specific heat of ice = C = 2100 J/(kg.K)

Latent heat of fusion for ice = L = 334 x 10³ J/kg

Mass of ice = m = 0.56 kg

Initial temperature of ice = T₁ = -19.6 ^oC

Melting point of ice = T₂ = 0 ^oC

Rate at which heat is supplied to the container = H = 760 J/min

Amount of time before the ice starts to melt = t₁

Ice will start to melt after all the ice reaches a temperature of 0 ^oC.

Heat to be supplied to the ice to reach a temperature of 0 ^oC = Q₁

Q₁ = mC(T₂ - T₁)

Q₁ = (0.56)(2100)(0 - (-19.6))

Q₁ = 23049.6 J

Q₁ = Ht₁

23049.6 = (760)t₁

t₁ = 30.33 minutes

Amount of time taken to increase the temperature of the ice above 0 ^oC = t₂

For the temperature to increase above 0 ^oC all the ice will need to reach 0 ^oC and all of it should melt.

Heat to be supplied to the ice to increase the temperature of the ice above 0 ^oC = Q₂

Q₂ = mC(T₂ - T₁) + mL

Q₂ = (0.56)(2100)(0 - (-19.6)) + (0.56)(334x10³)

Q₂ = 210089.6 J

Q₂ = Ht₂

210089.6 = (760)t₂

t₂ = 276.4 minutes

A) Amount of time before the ice starts to melt = 30.33 minutes

B) Amount of time taken to increase the temperature of the ice above 0 ^oC = 276.4 minutes