A kettle with 3 litre 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 $$= 4.2\,kJ/kg$$   )

Solution :
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.
$$\eqalign{ & Pt = ms\Delta \theta + {\text{energy}}\,{\text{lost}} \cr & 2000t = 3 \times 4.2 \times {10^3} \times \left( {97 - 27} \right) + 130t \cr & \Rightarrow t = 472\sec \cr} $$