Properties Of Solids And Liquids Question 436

Question: A kettle with 3 liter water at $ 27{}^\circ C $ is heated by operating coil heater of power 2 kW. The heat is lost to the atmosphere at constant rate 130 J/sec, when its lid is open. In how much time will water heated to $ 97{}^\circ C $ with the lid open? (specific heat of water =4.2kJ/kg)

Options:

A) 472 sec

B) 693 sec

C) 912 sec

D) 1101 sec

Show Answer

Answer:

Correct Answer: A

Solution:

[a] By the law of conservation of energy, energy given by heater must be equal to the sum of energy gained by water and energy lost from the lid. $ Pt=ms\Delta \theta + energy lost $

$ 2000t=3\times 4.2\times 10^{3}\times ( 97-27 )+130t $

$ \Rightarrow t=472sec $



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