Question

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

• A. 0.21
• B. 2.1
• C. 0.42
• D. 4.2

Hint:

In calorimetry problems, always ensure your units are consistent. Here, the specific heat is in J/kg, but the heat of combustion is in J/g. It's crucial to either convert everything to kg or everything to g. The question asks for the answer in g/s, so working with grams for the gas part is convenient.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to find the mass rate of gas consumption required to heat a certain volume flow rate of water by a specific temperature difference. This is a problem of energy conservation, where the heat energy released by the burning gas is absorbed by the water.
Step 2: Key Formula or Approach:
The rate of heat absorbed by water (Power absorbed) is given by:
\[ P_{\text{water}} = \dot{m}_{\text{water}} \cdot s \cdot \Delta T \] where \( \dot{m}_{\text{water}} \) is the mass flow rate of water, `s` is the specific heat capacity, and \( \Delta T \) is the temperature change.
The rate of heat supplied by the gas (Power supplied) is given by:
\[ P_{\text{gas}} = \dot{m}_{\text{gas}} \cdot L_c \] where \( \dot{m}_{\text{gas}} \) is the mass flow rate of gas and \( L_c \) is the heat of combustion.
By conservation of energy, \( P_{\text{water}} = P_{\text{gas}} \).
Step 3: Detailed Explanation:
1. Calculate the power absorbed by water:
Volume flow rate of water = 5.0 L/min.
Convert this to kg/s. The density of water is approximately 1 kg/L.
\[ \dot{m}_{\text{water}} = 5.0 \, \frac{\text{L}}{\text{min}} \times \frac{1 \, \text{kg}}{1 \, \text{L}} \times \frac{1 \, \text{min}}{60 \, \text{s}} = \frac{5}{60} \, \frac{\text{kg}}{\text{s}} = \frac{1}{12} \, \frac{\text{kg}}{\text{s}} \] Temperature change \( \Delta T = 87^\circ\text{C} - 27^\circ\text{C} = 60^\circ\text{C} \).
Specific heat of water \( s = 4200 \) J/kg.°C.
\[ P_{\text{water}} = \left( \frac{1}{12} \, \frac{\text{kg}}{\text{s}} \right) \times \left( 4200 \, \frac{\text{J}}{\text{kg}\cdot^\circ\text{C}} \right) \times (60^\circ\text{C}) \] \[ P_{\text{water}} = \frac{4200 \times 60}{12} = 4200 \times 5 = 21000 \, \frac{\text{J}}{\text{s}} = 21000 \, \text{W} \] 2. Calculate the rate of gas consumption:
Let the rate of gas consumption be \( \dot{m}_{\text{gas}} \) in g/s.
Heat of combustion \( L_c = 5.0 \times 10^4 \) J/g.
\[ P_{\text{gas}} = \dot{m}_{\text{gas}} \times (5.0 \times 10^4 \, \text{J/g}) \] Equating the power supplied and absorbed:
\[ P_{\text{gas}} = P_{\text{water}} \] \[ \dot{m}_{\text{gas}} \times 50000 = 21000 \] \[ \dot{m}_{\text{gas}} = \frac{21000}{50000} = \frac{21}{50} = \frac{42}{100} = 0.42 \, \text{g/s} \] Step 4: Final Answer:
The rate of consumption of the gas is 0.42 g/s. This corresponds to option (C).