Water flow rate = 5.0 litres per minute.
Since density of water ≈ 1 kg per litre,
Mass per minute = 5 kg/min.
Convert into kg/s:
\[ \dot{m} = \frac{5}{60} = 0.0833 \text{ kg/s} \]
Temperature rise:
\[ \Delta T = 87 - 27 = 60^\circ C \]
Heat required per second:
\[ \dot{Q} = \dot{m} c \Delta T \]
\[ \dot{Q} = 0.0833 \times 4200 \times 60 \]
\[ \dot{Q} = 21000 \text{ J/s} \]
Let rate of gas consumption be \(x\) g/s.
Heat supplied per second:
\[ x \times 5.0 \times 10^4 \]
Equating:
\[ x \times 5.0 \times 10^4 = 21000 \]
\[ x = 0.42 \text{ g/s} \]
Therefore correct answer is (D).
Calorimetry is a foundational topic in Thermal Physics and plays a crucial role in competitive examinations like JEE Main, JEE Advanced, and NEET. The core idea behind calorimetry is the principle of conservation of energy. Whenever heat is transferred from one body to another, the total energy remains conserved. Heat lost by one system equals heat gained by another, provided there is no loss to the surroundings.
The fundamental equation used in calorimetry is:
\[ Q = mc\Delta T \]
Here Q represents the heat absorbed or released, m is mass, c is specific heat capacity, and ΔT is change in temperature. Specific heat capacity is defined as the amount of heat required to raise the temperature of 1 kg of a substance by 1°C. Water has a very high specific heat capacity (4200 J/kg°C), which is why it is used extensively as a coolant in automobiles, nuclear reactors, and industrial heat exchangers.
In flow systems like geysers, boilers, and heat exchangers, we use rate form of calorimetry:
\[ \frac{dQ}{dt} = \dot{m} c \Delta T \]
Here \(\dot{m}\) represents mass flow rate in kg/s. In competitive exams, students often forget to convert litres per minute into kg per second. Since density of water ≈ 1000 kg/m³, 1 litre = 1 kg. Therefore, direct conversion is possible.
Heat of combustion is another extremely important concept. It is defined as the amount of heat released when 1 gram (or 1 kg) of a fuel burns completely in oxygen. It represents chemical energy stored in molecular bonds. When gas burns in a geyser, chemical energy converts into thermal energy which is transferred to water.
The principle applied in this question is:
\[ \text{Heat released by fuel per second} = \text{Heat absorbed by water per second} \]
This equality assumes 100% efficiency. In real life, efficiency may be 60–80% due to heat loss through exhaust gases and surroundings.
Core Concepts Involved:
• Specific heat capacity
• Heat of combustion
• Mass flow rate
• Energy conservation principle
• Unit consistency and dimensional correctness
Important Formulas for Exams:
\[ Q = mc\Delta T \]
\[ \dot{Q} = \dot{m}c\Delta T \]
\[ \text{Fuel energy} = (\text{mass rate}) \times (\text{heat of combustion}) \]
\[ \eta = \frac{\text{Useful heat output}}{\text{Heat supplied by fuel}} \]
Efficiency questions are common in JEE Advanced. Sometimes the problem gives efficiency and asks fuel consumption. In that case:
\[ \dot{m}_{fuel} = \frac{\dot{m}_{water} c \Delta T}{\eta \times H} \]
where H is heat of combustion.
Conceptual Understanding:
Heat is a form of energy transfer due to temperature difference. It is not a substance but a mode of energy transfer. In calorimetry, we assume no work is done and no phase change occurs unless specified.
If phase change is involved, then latent heat formula is used:
\[ Q = mL \]
where L is latent heat. JEE frequently mixes sensible heat (mcΔT) and latent heat (mL) in the same question.
Common Mistakes Students Make:
• Forgetting to convert minutes to seconds.
• Not converting litres to kilograms.
• Using incorrect specific heat values.
• Arithmetic mistakes in multiplication.
• Ignoring efficiency when mentioned.
Exam Relevance:
In JEE Main, 1–2 questions from Thermal Physics are common every year. Calorimetry problems are generally direct but require careful unit handling. In JEE Advanced, calorimetry may be combined with thermodynamics laws, entropy, or ideal gas processes.
Advanced Insight:
From thermodynamics perspective, this problem is based on First Law of Thermodynamics:
\[ \Delta Q = \Delta U + W \]
In a geyser heating water, no mechanical work is done by water. Hence almost all heat goes into increasing internal energy, which appears as temperature rise.
Continuous flow heating systems are important in engineering. Boilers in thermal power plants work on the same principle but at larger scales. Understanding these concepts builds bridge between school-level physics and engineering applications.
Real-Life Applications:
• Gas geysers
• Solar water heaters
• Industrial boilers
• Automobile radiators
• Nuclear reactor cooling systems
Mastering calorimetry improves numerical solving speed and builds confidence in Thermal Physics. Students should practice mixture problems, ice–water equilibrium, steam condensation and fuel efficiency problems to strengthen preparation.
This topic is fundamental for JEE Main, JEE Advanced, NEET and engineering entrance exams because it combines conceptual clarity with numerical accuracy.
1. Why is density of water taken as 1 kg/L?
Because at room temperature density of water is approximately 1000 kg/m³ which equals 1 kg per litre.
2. What if efficiency was given?
Then divide useful heat by efficiency to calculate required fuel energy.
3. Is calorimetry important for JEE Advanced?
Yes, especially in multi-concept thermodynamics problems.