Question

Water is boiled in a pan on a stove at sea level. During 10 min of boiling, it is observed that 200 g of water has evaporated. Then the rate of heat transfer to the water is \((a) 0.84 \mathrm{kJ} / \mathrm{min}\) \((b) 45.1 \mathrm{kJ} / \mathrm{min}\) \((c) 41.8 \mathrm{kJ} / \mathrm{min}\) \((d) 53.5 \mathrm{kJ} / \mathrm{min}\) \((e) 225.7 \mathrm{kJ} / \mathrm{min}\)

Short answer

Answer: (b) 45.1 kJ/min

Step-by-step solution

Identify the given values

- Mass of water evaporated (m) = 200 g - Time taken (t) = 10 min - Latent heat of vaporization of water (L) = 2.26 × 10^6 J/kg (given)

Conversion of variables to appropriate units

Convert the mass of water evaporated from grams to kilograms (since the latent heat is given per kg). m = 200 g × (1 kg / 1000 g) = 0.2 kg Convert the time taken from minutes to seconds (since the heat transfer rate will be in J/s). t = 10 min × (60 s / min) = 600 s

Calculate the total heat required to evaporate the water

Use the formula Heat = mass × evaporation heat capacity to find the total heat (Q) required to evaporate the given mass of water. Q = m × L = 0.2 kg × 2.26 × 10^6 J/kg = 4.52 × 10^5 J

Calculate the rate of heat transfer

Divide the total heat (Q) by the time (t) to find the rate of heat transfer (H). H = Q / t = 4.52 × 10^5 J / 600 s = 753.33 J/s Now, we need to match the result with one of the given options. We can convert 753.33 J/s to kJ/min by using the following conversions: 1 J = 0.001 kJ 1 s = 1/60 min H = 753.33 J/s × (0.001 kJ/J) × (60 s/min) = 45.2 kJ/min The closest option to this value is: (b) 45.1 kJ/min