Question

A stove consumes 1 gram of kerosene in 48 seconds. if the calorific value of kerosene is 48 KJ / gm, then the power of consumption of the stove in kW is

Answer 1

48 kJ / gram * 1 gram / 48 seconds = 1 kJ / second = 1 kW.

Answer 2

heya answer is here......


◇ 48 kj/gram × 1 gram/48 seconds= 1kj/second= 1KW


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