Thermodynamics Part 5

Concept:The heat required to raise water temperature is provided by burning gas, and under 100% efficiency, the rate of heat produced equals the rate of heat absorbed by water.Explanation:Water flow rate $= 5.0$ L/min. Density of water $= 1$ kg/L, so mass flow rate $= 5.0$ kg/min $= \frac{5.0}{60}$ kg/s. Temperature rise $\Delta T = 87^\circ\mathrm{C} - 27^\circ\mathrm{C} = 60^\circ\mathrm{C}$. Specific heat of water $c = 4200$ J/(kg·°C). Rate of heat required: $P_{\text{req}} = \frac{dm}{dt} \cdot c \cdot \Delta T$. $P_{\text{req}} = \left(\frac{5.0}{60}\right) \times 4200 \times 60$ J/s $= 21000$ J/s. Let gas consumption rate be $R$ g/s. Heat of combustion $H = 5.0 \times 10^4$ J/g. Rate of heat produced: $P_{\text{prod}} = R \times H$. Assuming 100% efficiency: $P_{\text{prod}} = P_{\text{req}}$. $R \times 5.0 \times 10^4 = 21000$. $R = \frac{21000}{50000} = 0.42$ g/s.Answer:$0.42$ g/s, which corresponds to option D.