First Law of Thermodynamics Problems on Internal Energy Rate Change

Heat transfer to a system and mechanical work output are central ideas in introductory thermodynamics.

This post deals with the first law of thermodynamics, specifically the bookkeeping of energy entering and leaving a system. Heat is supplied to a system while the system simultaneously performs work on its surroundings. The conceptual task is to determine the rate of increase of internal energy by balancing the energy inflow and outflow using a physically consistent sign convention.

In differential form, the governing relation is the first law: internal energy changes according to net energy transfer. When heat enters the system at rate ##\dot Q## and work is done by the system at rate ##\dot W##, the rate of change of internal energy is ##\frac{dU}{dt}=\dot Q-\dot W##. The source image therefore tests conceptual fluency rather than computational complexity.

Problem 1: Thermal Input and Shaft Work in a Closed Vessel

This problem uses the same underlying concept: the internal energy increase equals the heat supplied minus the work done by the system. In thermodynamics, a positive heat-transfer term raises system energy, whereas positive work done by the system represents energy leaving the system. Careful attention to signs prevents the most common errors in elementary energy-balance questions.

Such questions are scientifically important because they train students to distinguish between energy storage and energy transfer. Power, measured in watts, expresses energy change per unit time. A system may receive substantial heat and still store only part of it if some energy is simultaneously converted into mechanical work, electrical work, or boundary expansion against the surroundings.

Problem Statement

A gas in a rigidly insulated apparatus receives heat from an embedded coil at a rate of 180 W. At the same time, a paddle wheel attached to the gas delivers mechanical output so that the system does work at a rate of 65 W. Determine the rate at which the internal energy of the gas increases.

Problem Statement Explanation

The essential idea is to compare incoming and outgoing energy rates. Heat transfer to the system is an energy gain, while work performed by the system is an energy loss. Because the question asks for the rate of internal energy increase, we compute the net power retained inside the system rather than the total heat supplied alone.

Complete Detailed Solution

Using the thermodynamic rate form of the first law, we write the internal-energy balance as ##\frac{dU}{dt}=\dot Q-\dot W##. Here, the heat input is ##\dot Q=180\ \text{W}## and the work output is ##\dot W=65\ \text{W}##. Substituting the values gives the required rate of energy storage.

###\frac{dU}{dt}=180-65=115\ \text{W}###

Therefore, the internal energy of the gas increases at a rate of 115 W. The result is positive, which means the system stores energy overall. Even though it performs work externally, the incoming heat is larger than the work output, so the difference remains available to raise internal microscopic energy.

thermodynamic rule

Energy-Rate Sign Convention Summary

A compact guide to interpreting heat input and work output in closed-system power balances.

Note:

• Use one sign convention consistently throughout the calculation.

• Rate questions are expressed in watts, where ##1\ \text{W}=1\ \text{J s}^{-1}##.

Problem 2: Electrical Heating with Piston Expansion

The same first-law framework applies when the mode of work changes. Instead of shaft work, the system may do boundary work by expanding a piston. The conceptual structure remains identical: determine whether energy entering as heat exceeds energy leaving as work. If it does, internal energy rises; otherwise, it falls despite continued heating.

These exercises also reinforce the distinction between process mechanism and energy accounting. Whether the system performs electrical, shaft, or expansion work is secondary in a simple closed-system balance. At an introductory level, only the magnitudes and directions of heat transfer and work transfer matter for calculating the internal energy storage rate.

Problem Statement

A cylinder containing vapor receives heat at a rate of 220 W from a surrounding furnace. Simultaneously, the vapor expands and performs boundary work on a piston at a rate of 140 W. Find the rate of increase of the vapor’s internal energy.

Problem Statement Explanation

The furnace adds energy to the vapor, while piston expansion carries some of that energy out as work. The net difference between those two power terms is the power retained in the system. Because the problem asks for internal energy increase, we again use the closed-system rate equation without changing its mathematical form.

Complete Detailed Solution

Apply ##\frac{dU}{dt}=\dot Q-\dot W## with ##\dot Q=220\ \text{W}## and ##\dot W=140\ \text{W}##. Substituting gives the net rate of internal energy storage. Since the supplied heat is larger than the work output, the answer must be positive, consistent with an increase in microscopic energy.

###\frac{dU}{dt}=220-140=80\ \text{W}###

Hence, the internal energy of the vapor increases at 80 W. This means every second the system retains 80 J after exporting 140 J as work. The problem illustrates that a heated expanding system may still warm internally, but only at the reduced rate given by the net energy balance.

Problem 3: Heated Fluid with Strong Mechanical Output

A scientifically interesting variation occurs when work output is large compared with heat input. Students often assume heating automatically means internal energy must increase. However, the first law shows that if the system exports more energy as work than it receives as heat, the internal energy decreases, even though thermal energy is still being supplied continuously.

This is a useful conceptual checkpoint because it emphasizes conservation rather than intuition. Internal energy is not determined by heat transfer alone; it depends on the algebraic sum of all energy-transfer channels. Such reasoning underlies thermal machines, engines, and many laboratory devices where energy simultaneously enters and leaves through different physical mechanisms.

Problem Statement

A thermodynamic system receives heat at a rate of 95 W. During the same interval, it performs work on the surroundings at a rate of 130 W. Determine the rate of change of the system’s internal energy and state whether it is increasing or decreasing.

Problem Statement Explanation

This problem differs slightly because the work output exceeds the heat input. The mathematical procedure is unchanged, but the interpretation is important. A negative result for ##\frac{dU}{dt}## means the system is losing stored energy overall. That stored energy may be converted into work in addition to the incoming thermal supply.

Complete Detailed Solution

From the first law in rate form,

###\frac{dU}{dt}=\dot Q-\dot W###

Substitute ##\dot Q=95\ \text{W}## and ##\dot W=130\ \text{W}##:

###\frac{dU}{dt}=95-130=-35\ \text{W}###

The internal energy changes at ##-35\ \text{W}##, so it is decreasing at a rate of 35 W. Physically, the system delivers more power outward than it receives through heating. The 35 W deficit must come from its own stored internal energy, which therefore declines as the process continues.

comparison

Sample Energy-Balance Outcomes

Representative cases showing how relative magnitudes of heat and work determine the sign of internal-energy change.

Note:

• Only relative magnitudes are needed to predict the direction of change.

• This logic applies broadly to elementary closed-system thermodynamic processes.

Problem 4: Equal Heat Input and Work Output

Another instructive case occurs when the incoming heat rate and outgoing work rate are numerically equal. Students may incorrectly search for a hidden complication, yet the first law yields a clean result: no net energy is stored. Such balance conditions are pedagogically valuable because they highlight equilibrium-like energy flow without implying thermal or mechanical stasis.

In scientific modeling, a zero rate of internal-energy change does not mean nothing happens. Energy may continue flowing through the system continuously, with heat entering and work leaving at identical rates. The system then acts as an energy-throughput device, transmitting power without accumulating or depleting internal energy over the interval considered.

Problem Statement

A system absorbs heat at a rate of 150 W while simultaneously doing work on the surroundings at a rate of 150 W. Find the rate of change of the internal energy and interpret the result physically.

Problem Statement Explanation

Because the two energy-transfer rates are equal, their difference should vanish. The system is not isolated; rather, it is energetically balanced. The question therefore tests whether the student understands that internal energy depends on the net of all transfers, not on the presence of heat input considered alone.

Complete Detailed Solution

Use the first-law rate equation directly:

###\frac{dU}{dt}=\dot Q-\dot W=150-150=0\ \text{W}###

Thus, the rate of change of internal energy is zero. Physically, the system receives 150 J each second as heat and releases exactly 150 J each second as work. Since no net energy remains within the system, its internal energy stays constant during the process under the stated conditions.

Problem 5: Resistive Heating with Modest Work Delivery

This final variation preserves the same core principle but places it in an engineering context involving electrical heating. Resistive elements are common in laboratory and industrial devices. When a heated system also performs work, perhaps by moving a diaphragm or rotating a small rotor, the internal-energy rate must still be computed from the net power balance.

From an instructional standpoint, repetition across contexts improves transfer of learning. Once students understand that the first law is fundamentally an accounting statement, they can apply it to gases, liquids, biological systems, electrothermal devices, and mechanical assemblies. The surrounding story may change, but the governing conservation principle remains mathematically invariant.

Problem Statement

An electrically heated chamber receives thermal energy at a rate of 260 W. During operation, the chamber drives a mechanical linkage and performs work at a rate of 85 W. Calculate the rate at which the chamber’s internal energy increases.

Problem Statement Explanation

The chamber gains energy from heating and loses part of it through work. Since the heating term exceeds the work term, the internal energy must rise. The computation is a direct application of the same conservation statement used in all previous examples, demonstrating the unity of thermodynamic reasoning across different devices.

Complete Detailed Solution

Again, apply

###\frac{dU}{dt}=\dot Q-\dot W###

With ##\dot Q=260\ \text{W}## and ##\dot W=85\ \text{W}##, we obtain

###\frac{dU}{dt}=260-85=175\ \text{W}###

Therefore, the internal energy increases at a rate of 175 W. This indicates strong net energy storage within the chamber. The example also demonstrates how a system can simultaneously act as a heater-driven device and a work-producing device, with conservation laws quantifying the balance precisely and unambiguously.

dataset

Generated Problem Results Overview

Computed internal-energy rates for the five analogous thermodynamic scenarios developed from the source concept.

Note:

• Positive values denote internal-energy increase; negative values denote decrease.

• All examples use the same conservation equation with different numerical conditions.

The source image itself yields a straightforward answer. Heat enters at ##100\ \text{W}## and work leaves at ##75\ \text{J s}^{-1}=75\ \text{W}##. Therefore, by the first law, the internal energy increases at ##100-75=25\ \text{W}##. Hence the correct option is (c) 25 W, and the comprehensive meaning is conservation of energy in a thermodynamic system with simultaneous heat input and work output.