A gas based geyser heats water flowing at the rate of 5.0$5.0$ litres per minute from 27^circtextC$27^{\circ}\text{C}$ to 87^circtextC$87^{\circ}\text{C}$. The rate of consumption of the gas is ____ g/s. (Take heat of combustion of gas = 5.0 times 10^4$5.0 \times 10^{4}$ J/g, specific heat capacity of water = 4200text J/kg.^circtextC$4200\text{ J/kg.}^{\circ}\text{C}$)

### Related Formula P = fracdmdt S Delta T$$P = \frac{dm}{dt} S \Delta T$$ P_textheater = textRate of combustion times textHeat of combustion$$P_{\text{heater}} = \text{Rate of combustion} \times \text{Heat of combustion}$$ ### Core Logic Water flow rate = 5text l/min = frac560text kg/s = frac112text kg/s$5\text{ l/min} = \frac{5}{60}\text{ kg/s} = \frac{1}{12}\text{ kg/s}$. The power required to heat the water is: P_textheater = left(fracdmdtright)_textwater cdot S cdot Delta T$$P_{\text{heater}} = \left(\frac{dm}{dt}\right)_{\text{water}} \cdot S \cdot \Delta T$$ P_textheater = frac112 times 4200 times (87 - 27) = frac112 times 4200 times 60text W$$P_{\text{heater}} = \frac{1}{12} \times 4200 \times (87 - 27) = \frac{1}{12} \times 4200 \times 60\text{ W}$$ ### Step 1: Solving for Gas Consumption Let the rate of consumption of gas be xtext g/s$x\text{ g/s}$. Heat generated by the gas per second must equal the power required to heat the water: x times 5.0 times 10^4 = frac112 times 4200 times 60$$x \times 5.0 \times 10^{4} = \frac{1}{12} \times 4200 \times 60$$ x times 5.0 times 10^4 = 4200 times 5$$x \times 5.0 \times 10^{4} = 4200 \times 5$$ x = frac4200 times 55 times 10^4 = 4200 times 10^-4$$x = \frac{4200 \times 5}{5 \times 10^{4}} = 4200 \times 10^{-4}$$ x = 0.42text g/s$$x = 0.42\text{ g/s}$$

### Pattern Recognition Equate the thermal power absorbed by fluid (dm/dt cdot c cdot Delta T$dm/dt \cdot c \cdot \Delta T$) with the power output of fuel (dm_textfuel/dt cdot L_textcombustion$dm_{\text{fuel}}/dt \cdot L_{\text{combustion}}$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermal Properties of Matter