Question 1 of 7intermediate🔧 ApplyNumerical3 marks

A geyser heats water flowing at the rate of 3.0 litres per minute from 27 °C to 77 °C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is 4.0×104 J/g4.0 \times 10^4 \text{ J/g}?

Correct Answer

15.75 g/min

Exercise: EXERCISES | Q: 11.1 | (Page Page 17)
For More Understanding

Explanation

To find the rate of fuel consumption, we first determine the amount of heat energy required to heat the water per minute. This is calculated using the heat capacity formula provided in the context (S=ΔQ/ΔTS = \Delta Q / \Delta T), rearranged to Q=msΔTQ = ms\Delta T. We use standard values for the specific heat capacity of water (4.2 J g−1 K−14.2 \text{ J g}^{-1} \text{ K}^{-1}) and the density of water (1 kg/L1 \text{ kg/L}), which are standard knowledge for Class 11 Physics problems involving water. Once the total heat required per minute is found, we divide it by the heat of combustion of the fuel to find the mass of fuel consumed per minute.

Solution Steps

  1. Step 1: Calculate the mass of water heated per minute. The volume flow rate is 3.0 litres/minute. Using the standard density of water (1 kg/L1 \text{ kg/L}), the mass is m=3.0 kg=3000 gm = 3.0 \text{ kg} = 3000 \text{ g}.

  2. Step 2: Calculate the heat required to raise the temperature. The temperature change is ΔT=77∘C−27∘C=50∘C\Delta T = 77^\circ\text{C} - 27^\circ\text{C} = 50^\circ\text{C}. Using the specific heat capacity of water s=4.2 J g−1 K−1s = 4.2 \text{ J g}^{-1} \text{ K}^{-1}, the heat supplied is Q=msΔT=3000 g×4.2 J g−1 K−1×50 K=630,000 JQ = ms\Delta T = 3000 \text{ g} \times 4.2 \text{ J g}^{-1} \text{ K}^{-1} \times 50 \text{ K} = 630,000 \text{ J}.

  3. Step 3: Calculate the rate of fuel consumption. The heat of combustion is given as 4.0×104 J/g4.0 \times 10^4 \text{ J/g}. The mass of fuel required per minute is Mass=Total HeatHeat of combustion=630,000 J40,000 J/g=15.75 g\text{Mass} = \frac{\text{Total Heat}}{\text{Heat of combustion}} = \frac{630,000 \text{ J}}{40,000 \text{ J/g}} = 15.75 \text{ g}.

  4. Step 4: Since the water flow rate is per minute, the rate of fuel consumption is 15.75 g/min.

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