Question

A pot is being used to boil off 1 kg of water. The specific energy required to cause the phase change is 2297 kJ/kg. Assuming the stovetop supplies 20 kJ/s to the water and the liquid is at boiling temperature, how long will it take to vaporize half of the water? Report your answer in seconds to the nearest whole number don't knou 2 attemots

Answer 1

Answer:

58 seconds

Step-by-step explanation:

Given:

Initial mass of water = 1 kg

Specific energy = 2297 kJ/kg

Heat supplied by the stove = 20 kJ/s

Now,

Half water is to be vaporized i.e 0.5 kg

Thus, heat required for vaporizing 0.5 kg water = mass × specific heat

or

heat required for vaporizing 0.5 kg water = 0.5 × 2297 = 1148.5 kJ

Therefore,

time taken to provide the required heat = $\frac{\textup{Heat required}}{\textup{Heat supplied per second}}$

or

time taken to provide the required heat = $\frac{\textup{1148.5 kJ}}{\textup{20 kJ/s}}$

or

time taken to provide the required heat = 57.425 ≈ 58 seconds