Answer :
Final answer:
Approximately 100 grams of water would boil away if a 1.25 kW microwave oven operates for 3 minutes, but this assumes all energy is used for vaporizing water, which is not typically the case in practice.
The correct option is a
Explanation:
To determine how many grams of water will boil away if the oven is operated for 3 minutes, we need to calculate the energy transferred from the oven to the water and then use the heat of vaporization to find the mass of the evaporated water.
Firstly, the power of the microwave oven is 1.25 kW, which is equivalent to 1250 W (Watts). Power is defined as energy per unit time (P = E/t). Thus, operating the oven for 3 minutes (180 seconds) at 1.25 kW means the total energy provided by the oven is:
E (total energy) = Power × Time
E = 1250 W × 180 s = 225000 J (Joules)
Given that the heat of vaporization for water is approximately 2260 kJ/kg (or 2260000 J/kg), we can determine the mass of water (m) that can be boiled away as follows:
m = E / heat of vaporization
m = 225000 J / 2260000 J/kg = 0.09956 kg
Converting the mass to grams (1 kg = 1000 g):
m = 0.09956 kg × 1000 g/kg ≈ 99.56 g
This calculation suggests that if the microwave operates for 3 minutes, approximately 100 grams of water would boil away, meaning option (a) 150 grams would be the closest estimate, but actual data suggests less than that amount would evaporate. We must have additional information regarding the efficiency or if all energy goes into vaporizing the water to provide a precise answer.
The correct option is a
