Question

An open container holds ice of mass 0.550 kg at a temperature of -11.4 ∘C . The mass of the container can be ignored. Heat is supplied to the container at the constant rate of 890 J/minute . The specific heat of ice to is 2100 J/kg⋅K and the heat of fusion for ice is 334×103J/kg.

Part A How much time tmelts passes before the ice starts to melt? View Available Hint(s) tmelts tmeltst_melts =   minutes   SubmitPrevious Answers Incorrect; Try Again; 7 attempts remaining Part B From the time when the heating begins, how much time trise does it take before the temperature begins to rise above 0∘C? View Available Hint(s) trise triset_rise =   minutes   SubmitPrevious Answers Incorrect; Try Again; 7 attempts remaining

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 = 0.55 kg

Initial temperature of ice = T₁ = -11.4 ^oC

Melting point of ice = T₂ = 0 ^oC

Rate at which heat is supplied to the container = H = 890 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 = (0.55)(2100)(0 - (-11.4))

Q₁ = 13167 J

Q₁ = Ht₁

13167 = (890)t₁

t₁ = 14.79 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 = (0.55)(2100)(0 - (-11.4)) + (0.55)(334x10³)

Q₂ = 196867 J

Q₂ = Ht₂

196867 = (890)t₂

t₂ = 221.2 minutes

A) Amount of time before the ice starts to melt = 14.79 minutes

B) Amount of time taken to increase the temperature of the ice above 0 ^oC = 221.2 minutes