Answer:
58 seconds
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}}](https://img.qammunity.org/2020/formulas/mathematics/college/ovp3q1tvqzl31ohbnkfmkhrti3fc3dwe39.png)
or
time taken to provide the required heat =
![\frac{\textup{1148.5 kJ}}{\textup{20 kJ/s}}](https://img.qammunity.org/2020/formulas/mathematics/college/anyj9g2vuklri1gmjde8t505824kxwklki.png)
or
time taken to provide the required heat = 57.425 ≈ 58 seconds