# The First Law of Thermodynamics: Top 5 Numerical Problems and Solutions

### Abstract:

The First Law of Thermodynamics, a fundamental principle in physics and engineering, states that energy is conserved in any thermodynamic process. This article presents the top five numerical problems related to **the First Law of Thermodynamics**, offering step-by-step solutions for each scenario. These problems cover various aspects, ** including heat transfer**,

*,*

**work done****, and**

*internal energy changes***. By understanding and solving these numerical challenges, readers will gain a deeper insight into the practical applications of the First Law.**

*efficiency calculations*Top 5 Numerical problems and solutions of Thermodynamics |

### 1. Problem: Heat Transfer and Work Done in a Closed System

*Description:*

Consider a closed system containing 2 kg of an ideal gas undergoing a process. During this process, 1500 kJ of heat is transferred into the system, and the gas performs 500 kJ of work on its surroundings. Determine the change in internal energy of the gas and whether it is a heat addition or heat rejection process.

*Solution:*

The change in internal energy (Î”U) of the gas can be calculated using the ** First Law of Thermodynamics**:

**Î”U = Q - W**

Given that **Q = 1500 kJ** (heat transferred into the system) and **W = -500 kJ** (work done by the gas on its surroundings, as it is negative for work done on the surroundings),

**Î”U = 1500 kJ - (-500 kJ) = 1500 kJ + 500 kJ = 2000 kJ**

Since ** Î”U is positive (2000 kJ)**, the change in internal energy of the gas is

**positive**, indicating a heat addition process.

### 2. Problem: Efficiency of a Heat Engine

*Description:*

A heat engine takes in 8000 kJ of heat from a high-temperature reservoir and exhausts 4000 kJ of heat to a low-temperature reservoir while performing 2000 kJ of work. Calculate the efficiency of the heat engine.

*Solution:*

**The efficiency (Î·) of a heat engine** can be determined using the formula:

**Î· = (Work output / Heat input) * 100%**

Given that the work output is 2000 kJ, and the heat input is the sum of the heat absorbed from the high-temperature reservoir and the heat rejected to the low-temperature reservoir (8000 kJ + 4000 kJ = 12000 kJ), we can calculate:

**Î· = (2000 kJ / 12000 kJ) * 100% ≈ 16.67%**

The efficiency of the heat engine is approximately 16.67%.

### 3. Problem: Adiabatic Compression of an Ideal Gas

*Description:*

A cylinder contains 4 m³ of air at an initial pressure of 100 kPa and an initial temperature of 27°C. The air is compressed adiabatically until its volume reduces to 2 m³. Calculate the final pressure and temperature of the air.

*Solution:*

For an adiabatic process, we can use the relationship between pressure and volume given by:

**P₁ * V₁^Î³ = P₂ * V₂^Î³**

where ** P₁ and V₁ are the initial pressure and volume**,

**, and**

*P₂ and V₂ are the final pressure and volume***.**

*Î³ is the heat capacity ratio of the gas*First, we need to convert the initial temperature to Kelvin:

**T₁ = 27°C + 273.15 = 300.15 K**

Next, we calculate the heat capacity ratio (Î³) for air:

**Air is a diatomic gas, and for diatomic gases, Î³ ≈ 1.4.**

Now, we can solve for the final pressure (P₂):

**100 kPa * 4 m³^1.4 = P₂ * 2 m³^1.4**

**P₂ ≈ 317.07 kPa**

To find the ** final temperature (T₂)**, we can use the ideal gas law:

**P₂ * V₂ = n * R * T₂**

where n is the number of moles of air and R is the gas constant.

Since the process is adiabatic, there is no heat exchange, and the number of moles (n) remains constant. Therefore:

**P₁ * V₁ / T₁ = P₂ * V₂ / T₂**

Solving for **T₂**:

**T₂ = (P₂ * V₂ * T₁) / (P₁ * V₁)**

**T₂ = (317.07 kPa * 2 m³ * 300.15 K) / (100 kPa * 4 m³)**

**T₂ ≈ 475.72 K (approx)**

### 4. Problem: Energy Balance in a Steam Turbine

*Description:*

In a steam turbine, steam enters at 5 MPa and 400°C with a mass flow rate of 10 kg/s. The steam exits at 0.2 MPa and 150°C. Calculate the power output of the turbine assuming steady-state operation and neglecting kinetic and potential energy changes.

*Solution:*

The power output of the steam turbine can be calculated using the ** energy balance equation**:

**Power output = Mass flow rate * (h₁ - h₂)**

where * h₁ and h₂ are the specific enthalpies of the steam at the inlet and outlet*, respectively.

To find h₁ and h₂, we need to use the steam tables or steam properties calculator to look up the specific enthalpies at the given conditions. Assuming the values are:

**h₁ ≈ 3350 kJ/kg (specific enthalpy at 5 MPa and 400°C)**

**h₂ ≈ 2850 kJ/kg (specific enthalpy at 0.2 MPa and 150°C)**

Now, we can calculate the ** power output**:

**Power output = 10 kg/s * (3350 kJ/kg - 2850 kJ/kg) = 5000 kW**

The power output of the steam turbine is 5000 kW.

### 5. Problem: Efficiency of an Energy Storage System

*Description:*

An energy storage system stores 5000 kJ of electrical energy and delivers 4200 kJ of usable energy. Determine the efficiency of the energy storage system.

*Solution:*

The efficiency (**Î·_storage**) of an energy storage system can be calculated using the formula:

**Î·_storage = (Energy output / Energy input) * 100%**

Given that ** the energy input is 5000 kJ** and

**, we can calculate:**

*the energy output (usable energy) is 4200 kJ***Î·_storage = (4200 kJ / 5000 kJ) * 100% ≈ 84%**

The efficiency of the energy storage system is approximately 84%.

### Conclusion:

* The First Law of Thermodynamics plays a critical role in understanding* and

**analyzing various numerical problems in thermodynamics and energy systems**. By applying the principles of energy conservation and utilizing appropriate equations, engineers and scientists can solve complex numerical challenges related to

**,**

*heat transfer***,**

*work done***,**

*internal energy changes***, and more. Understanding these numerical problems and their solutions is essential for**

*efficiency***and**

*the effective design***, ultimately leading to more efficient and sustainable energy utilization.**

*optimization of thermodynamic systems*