A cup containing 100g of pure water at 20°C is placed in a refrigerator. If the refrigerator extracts heat at a rate of 840J per minute, calculate the time ...

Answer Details

To freeze water, it must be cooled from its initial temperature of 20°C to 0°C and then undergo a phase change from liquid to solid at 0°C. The amount of heat extracted from the water during cooling is given by: Q1 = mcΔT Where Q1 is the amount of heat, m is the mass of the water, c is its specific heat capacity, and ΔT is the change in temperature. Substituting the given values: Q1 = (100g) x (4.2J/gK) x (20°C-0°C) = 8,400J The amount of heat required for the water to freeze is given by: Q2 = mL Where Q2 is the amount of heat, m is the mass of the water, and L is the specific latent heat of fusion. Substituting the given values: Q2 = (100g) x (336J/g) = 33,600J The total amount of heat that must be extracted from the water to freeze it is the sum of Q1 and Q2: Qtot = Q1 + Q2 = 8,400J + 33,600J = 42,000J The rate at which heat is extracted from the water is given as 840J/min. Therefore, the time required to extract the total amount of heat required for freezing is: t = Qtot / heat extraction rate = 42,000J / 840J/min = 50 minutes Therefore, the time taken for the water to freeze is 50 minutes. Answer: 4