Question

2 litre water at 27°C is heated by a 1 kW heater in
an open container. On an average heat is lost to
surroundings at the rate 160 J/s. The time required
for the temperature to reach 77°C is
(1) 8 min 20 sec
(2) 10 min
(3) 7 min
(4) 14 min​

Answer 1

Answer:

The correct option is (1):  8 min 20 sec

Explanation:

From question heat gained by water in 1 second = 1 KW - 160 J/s = 1000 J/s - 160 J/s = 840 J/s

Total heat required to raise the temperature of water(volume 2L) from 27°C to 77°C

= m(water) x sp.ht x Δ θ

= 2 x 10^3 x 4.2 x 50  [ Q mass = density x volume]

Thus

840 x t =  2 x 10^3 x 4.2 x 50

OR

t =  2 x 10^3 x 4.2 x 50/840

t = 500 seconds

t = 8 min 20 sec

Answer 2

Answer:

8 min 20 sec

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 Δθ + energy lost

1000t=2×4.2×10^3×50+160t

840t=8.4×10^3 ×50=500sec

=8min 20sec