Question

Water is boiling at 1 atm pressure in a stainless steel pan on an electric range. It is observed that 2 kg of liquid water evaporates in 30 min. The rate of heat transfer to the water is \((a) 2.51 \mathrm{kW}\) (b) \(2.32 \mathrm{kW}\) \((c) 2.97 \mathrm{kW}\) \((d) 0.47 \mathrm{kW}\) \((e) 3.12 \mathrm{kW}\)

Short answer

Answer: The rate of heat transfer to the water is 2.51 kW.

Step-by-step solution

Find the heat of vaporization for water at 1 atm

The heat of vaporization for water at 1 atm pressure is approximately 40.7 kJ/mol.

Convert the mass of water to moles

To convert the given mass of water (2 kg) to moles, we will use the molar mass of water (18.015 g/mol). First, convert the mass of water to grams: $$2 \text{ kg} \times 1000 \frac{\text g}{\text{kg}} = 2000 \text g$$ Now, convert the mass to moles: $$\frac{2000 \text g}{18.015 \frac{\text g}{\text{mol}}} = 111.1 \text{ mol}$$

Calculate the total energy required for evaporation

Using the heat of vaporization and the number of moles, we can calculate the total energy required for the evaporation of 2 kg of water: $$111.1 \text{ mol} \times 40.7 \frac{\text{kJ}}{\text{mol}} = 4521.77 \text{ kJ}$$

Convert the time from minutes to seconds

We are given that the evaporation process takes 30 minutes. To determine the rate of heat transfer, we need to convert this time to seconds: $$30 \text{ min} \times 60 \frac{\text s}{\text{min}} = 1800 \text s$$

Calculate the rate of heat transfer

Finally, we can calculate the rate of heat transfer by dividing the total energy required by the time it took for evaporation: $$\frac{4521.77 \text{ kJ}}{1800 \text s} = 2.51 \frac{\text{kJ/s}}{\text{s}} = 2.51 \text{ kW}$$ The rate of heat transfer to the water is (a) \(2.51 \mathrm{kW}\).