Q.
A geyser heats water flowing at the rate of 3 litres per minute from 27∘C to 77∘C. If the geyser operates on a gas burner, and its heat of combustion is 4×104J/g, then the rate of consumption of fuel is
1 litre of water equals 1000g of water. ΔQ=msΔT (ΔtΔQ)=(Δtm)sΔT
As, swater =4.2Jg−1∘C−1 (ΔtΔQ)=1min(3000g)(4.2Jg−1∘C−1)(77−27)∘C (ΔtΔQ)=630000Jmin−1
As heat of combustion (L) L=(ΔmΔQ)=4×104Jg−1 ⇒(L)(ΔtΔm)=630000Jmin−1 ⇒ΔtΔm=4×104Jg−1630000Jmin−1≈16gmin−1