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 $
