Question

How much work is required, using an ideal Carnot refrigerator, to change 0.8 kg of tap water at 16⁰C into ice at −10⁰C ? Assume that the freezer compartment is held at −10⁰C and that the refrigerator exhausts energy into a room at 18⁰C. Take Cₐₜₑᵣ=4,186 J/kg⁰C,Cᵢₑ =2,090 J/kg⁰C,L=333,000 J/kg, and absolute zero to be at −273⁰C. [Give your answer in J to the nearest 1 J ]

Answer 1

To convert 0.8 kg of tap water at 16°C to ice at −10°C using an ideal Carnot refrigerator, the work required is approximately 21,758 J, after calculating the total heat removed from the water during cooling, freezing, and further cooling.

Step-by-step explanation:

To calculate the amount of work required to change 0.8 kg of tap water at 16°C into ice at −10°C using a Carnot refrigerator, we need to consider three processes:

• Cooling the water from 16°C to 0°C
• Freezing the water at 0°C
• Further cooling the ice from 0°C to −10°C

The total heat removed (Qc) is the sum of the heat removed in each process:

1. Q1 = m⋅Cwater⋅ΔT = 0.8 kg ⋅ 4186 J/kg°C ⋅ (16°C) = 53,491.2 J
2. Q2 = mL = 0.8 kg ⋅ 333,000 J/kg = 266,400 J
3. Q3 = m⋅Cice⋅ΔT = 0.8 kg ⋅ 2090 J/kg°C ⋅ (10°C) = 16,720 J

The total heat removed is Qc = Q1 + Q2 + Q3 = 336,611.2 J.

Using the Carnot refrigerator's coefficient of performance (COP), the work (W) required is:

W = Qc/COP, where COP = Tc/(Th - Tc).

The temperatures must be converted to Kelvin: Tc = 263 K (-10°C + 273), Th = 291 K (18°C + 273).

COP = 263/(291 - 263) = 15.47.

The work required is then: W = 336,611.2 J / 15.47 ≈ 21,758 J (to the nearest Joule)