Question -
Answer -
Given
Water is flowing at a rate of 3.0 litre/min
The geyser heats the water, raising the temperature from 270 C to 770 C.
Initial temperature, T1 = 270 C
Final temperature, T2 = 770 C
Rise in temperature, T = T2 – T1
= 77 – 27
= 500 C
Heat of combustion = 4 x 104 J/ g
Specific heat of water, C = 4.2 J / g 0C
Mass of flowing water, m = 3.0 litre / min
= 3000 g / min
Total heat used, Q = mcT
= 3000 x 4.2 x 50
On calculation, we get,
= 6.3 x 105 J/ min
Rate of consumption = 6.3 x 105 / (4 x104)
We get,
= 15.75 g/min
Therefore, rate of consumption is 15.75 g/min