Question

water of volume 2litre in a container is heated with a coil of 1kw at 27 degree celcios. the lid of the container is open and energy dissipates at rate of 106j/s.how much time temperature will rise from 27 degree celcios to 77 degree celcios (given specific heat of water =4.2j/kg) ​

Answer 1

answer : the net heat absorb by the water raise its temprature

=[1000-160]

=840 J/ s

now temprature raised till 77 clecius

so;

Q=mc delta t

2×4200 × 50J

t = Q/840 = 2× 4200× 50 / 840

= 500 sec = 8min and 20 s