Question

A geyser heats water flowing in it at the rate of 50 cc s^-1 from 27°C to 77°C. The geyser operates on a gas burner. It consumes 0.6 kg of gas per hour. The heat of combustion of gas is

(1) 6.3 x 10⁴ kJ(kg)^-1

(2) 6.3 x 10⁴ J(kg)^-1

(3) 6.3 x 10³ kJ(kg)^-1

(4) 6.3 x 10² kJ(kg)^-1

Answer 1

(1) 6.3 x 10⁴ kJ(kg)^-1

Heat absorbed by water flowing in 1s

= 50 x 1 x 1 x (77–27) Cal. 

= 2500 Cal. = 2500 x 4.2 J = 1.05 x 10⁴ J

Total heat absorbed in one hour = 1.05 x 10⁴ x 3600 

= 37.8 x 10⁶ J .....(1)

Let α be the heat of combustion of fuel in joule per kg.

Heat liberated by fuel in joule in one hour = 0.6 x α .....(2)

From Eqns (1) and (2) we have

37.8 x 10⁶ = 0.6α

\(∴\alpha=\frac{37.8}{0.6}\times10^6J\,kg^{-1}\)

= 6.3 x 10⁷ J kg^-1 = 6.3 x 10⁴ kJ kg^-1