Water of volume 2 L in a closed container is heated with a coil of 1 kW. While water is heated, the container loses energy at a rate of 160 J/s. In how much time will the temperature of water rise from 27 o C to 77 o C ? (Specific heat of water is 4.2 kJ/kg and that of the container is negligible).

Step-by-step solution

P = 1 0 0 0  watt ∴    P in = 1 0 0 0 - 1 6 0   J/s = 8 4 0  J/s Now, d Q dt = ms dT dt ∴     8 4 0 = ms dT dt ∴    dt = ms 8 4 0 dT Integrating both sides we have; t = ms 8 4 0 ∫ dT = 2 × 1 0 0 0 × 1 0 - 6 × 1 0 0 0 × 4.2 × 1 0 0 0 8 4 0 × 5 0 = 5 0 0 sec = 8  min 2 0 sec