Two litre of water at initial temperature of 27° C is heated by a heater of power 1 kW. If the lid of kettle is opened, then heat is lost at the constant rate of 160 J / s. Find, the time required to raise the temperature of water to 77° C with the lid open (specific heat of water 4.2 kJ / kg )

Question

Two litre of water at initial temperature of 27° C is heated by a heater of power 1 kW. If the lid of kettle is opened, then heat is lost at the constant rate of 160 J / s. Find, the time required to raise the temperature of water to 77° C with the lid open (specific heat of water 4.2 kJ / kg ) (A) 5 min 40 s (B) 14 min 20 s (C) 8 min 20 s (D) 16 min 10 s

Tardigrade Admin · Accepted Answer

Rate of heat gain =1000-160=840 J / s ∴ Required time =(2 × 4.2 × 103 ×(77-27)/840) =500 s =8 min 20 s

Q. Two litre of water at initial temperature of 27∘C is heated by a heater of power 1kW. If the lid of kettle is opened, then heat is lost at the constant rate of 160J/s. Find, the time required to raise the temperature of water to 77∘C with the lid open (specific heat of water 4.2kJ/kg )

5 min 40 s

14 min 20 s

8 min 20 s

16 min 10 s

Rate of heat gain =1000−160=840J/s ∴ Required time =8402×4.2×103×(77−27)​ =500s=8min20s

Rate of heat gain $= 1000 - 160 = 840 J / s$
$∴$ Required time $= \frac{2 \times 4.2 \times 1 0 ^{3} \times ( 77 - 27 )}{840}$
$= 500 s = 8 min 20 s$