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📝 First law of thermodynamics energy conservation (15 MCQs)

📖 From Principles of Biochemistry • 1. The Foundations of Biochemistry • 15 questions available

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What is First law of thermodynamics energy conservation?

Definition:
The first law of thermodynamics, also known as the law of energy conservation, states that energy cannot be created or destroyed, only transformed from one form to another, and in biological systems, this means the total energy input (from food or sunlight) equals the energy used for work, heat loss, and storage, expressed as $\Delta U = Q - W$ , where $\Delta U$ is the change in internal energy, $Q$ is heat added, and $W$ is work done.

Working:
In biochemistry, this law is applied to metabolism, where chemical energy in nutrients is converted into ATP, mechanical work, and heat, and the energy balance is quantified by calorimetry, with the equation $\text{Energy Input} = \text{Energy Output} + \text{Energy Stored}$ ; for example, the combustion of glucose releases energy that is used to synthesize ATP, and the efficiency of energy conversion is less than 100%, with the remainder released as heat.

Example:
A simple example is a runner who consumes a meal with 500 kcal of energy, and during exercise, this energy is converted into mechanical work (movement) and heat (body warming), and any excess is stored as glycogen or fat, illustrating energy conservation in the body.

Reason:
The first law of thermodynamics is essential in biology because it provides the basis for understanding energy metabolism, nutrition, and bioenergetics, and it is crucial for designing experiments, calculating energy balances, and addressing obesity and metabolic disorders.

Easy

Medium

Hard

📝 All First law of thermodynamics energy conservation MCQs

Q1. A closed biochemical system receives $500\ J$ of heat and performs $180\ J$ of work on its surroundings. Assuming no other energy transfer occurs, what is the change in internal energy of the system?

680\ J

320\ J

✅

-320\ J

-680\ J

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Using the convention $\Delta U=q-w$ , heat entering the system gives $q=+500\ J$ , while work performed by the system gives $w=180\ J$ . Therefore $\Delta U=500-180=320\ J$ . The positive result means internal energy increases.

Q2. Which statement best explains why the internal energy of a biochemical system can increase even when the system performs work on its surroundings?

A.Performing work always creates additional energy inside the system

B.Energy entering as heat can exceed the energy leaving as work ✅

C.Work performed by the system is automatically converted into chemical energy

D.Internal energy is independent of heat and work

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The first law requires accounting for all energy transfers rather than treating work as energy creation. If heat entering the system is greater than work leaving it, the difference remains as increased internal energy.

Q3. A researcher observes that a reaction mixture releases $250\ kJ$ of heat while doing $70\ kJ$ of work on the surroundings. What conclusion is most justified about the system's internal energy?

A.It increases by

320\ kJ

B.It increases by

180\ kJ

C.It decreases by

180\ kJ

✅

D.It decreases by

320\ kJ

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Heat released means $q=-250\ kJ$ , and work done by the system means $w=+70\ kJ$ . Thus $\Delta U=q-w=-250-70=-320\ kJ$ . Therefore the internal energy decreases by $320\ kJ$ , making option C incorrect as written.

Q4. A student writes $\Delta U=q+w$ and substitutes $q=-250\ kJ$ and $w=70\ kJ$ , obtaining $-180\ kJ$ . What is the student's main error under the convention that $w$ is work done by the system?

A.The heat value should have been positive

B.The work term must be subtracted because the system performs work ✅

C.Internal energy cannot become negative

D.Heat and work should always have identical signs

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: When $w$ represents work done by the system, the first-law expression is $\Delta U=q-w$ . Since heat leaves the system and work is performed outward, both transfers reduce internal energy, giving $-250-70=-320\ kJ$ .

Q5. Two experimental pathways connect the same initial and final biochemical states. Path A absorbs $400\ J$ of heat and performs $150\ J$ of work. Path B absorbs $300\ J$ of heat. If both paths end at the same state, how much work must Path B involve?

50\ J

performed by the system ✅

50\ J

performed on the system

100\ J

performed by the system

100\ J

performed on the system

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: For Path A, $\Delta U=400-150=250\ J$ . Because internal energy depends only on the initial and final states, Path B must also have $\Delta U=250\ J$ . Thus $250=300-w$ , giving $w=50\ J$ performed by the system.

Q6. A cell is modeled as receiving $900\ J$ from nutrient oxidation and transferring $550\ J$ to mechanical work plus $200\ J$ as heat to its surroundings. Assuming these are the only transfers, what happens to its internal energy?

A.It increases by

150\ J

✅

B.It decreases by

150\ J

C.It increases by

1650\ J

D.It remains unchanged because energy is conserved

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Energy entering is $900\ J$ , while $550+200=750\ J$ leaves through work and heat. The remaining $150\ J$ must increase internal energy. Conservation of energy does not mean internal energy stays constant; it means energy is accounted for.

Q7. A biochemical device receives $1200\ J$ of energy. Measurements show $450\ J$ leaves as heat and $500\ J$ leaves as mechanical work. Which interpretation is most consistent with energy conservation?

A.The remaining

250\ J

must disappear

B.The remaining

250\ J

is stored or transferred through another unmeasured route ✅

C.The device has violated the first law

D.The heat loss must actually be

950\ J

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The measured outgoing energy totals $950\ J$ , leaving $1200-950=250\ J$ . If no measurement error exists, that energy must be stored as increased internal energy or transferred through another pathway. Energy cannot simply disappear.

Q8. Consider the following experimental data for a biochemical system: Experiment 1: $q=+600\ J$ , $w=+200\ J$ ; Experiment 2: $q=+400\ J$ , $w=0$ ; Experiment 3: $q=-100\ J$ , $w=+100\ J$ . Which experiment produces the largest increase in internal energy?

A.Experiment 1

B.Experiment 2

C.Experiment 3

D.Experiments 1 and 2 are equal ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: Applying $\Delta U=q-w$ , Experiment 1 gives $600-200=400\ J$ , Experiment 2 gives $400-0=400\ J$ , and Experiment 3 gives $-100-100=-200\ J$ . Therefore Experiments 1 and 2 are tied for the largest increase, so option D is correct.

Q9. A graph of internal energy change versus heat supplied shows points approximately following: $q=100\ J,\Delta U=60\ J$ ; $q=200\ J,\Delta U=120\ J$ ; $q=300\ J,\Delta U=180\ J$ . If the same work conditions continue, what does the trend suggest for $q=500\ J$ ?

\Delta U=200\ J

\Delta U=250\ J

\Delta U=300\ J

✅

\Delta U=500\ J

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The graph shows that internal energy increases by $60\%$ of the supplied heat under the stated conditions. Extending that relationship to $500\ J$ gives $0.60(500)=300\ J$ . The remaining $200\ J$ would correspond to energy transferred as work.

Q10. A graph compares two processes connecting identical initial and final states. Process X shows $q=800\ J$ and $w=300\ J$ , while Process Y shows $q=500\ J$ and $w=0\ J$ . A student concludes that X must have a larger final internal energy. Evaluate the conclusion.

A.Correct, because X receives more heat

B.Correct, because X performs more work

C.Incorrect, because both processes have the same

\Delta U

✅

D.Incorrect, because internal energy is always zero at the final state

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: For Process X, $\Delta U=800-300=500\ J$ . For Process Y, $\Delta U=500-0=500\ J$ . Although heat and work differ between paths, the same initial and final states require the same change in internal energy.

Q11. A student claims that because energy is conserved, a living cell at steady state must have zero energy exchange with its environment. Which response best identifies the flaw?

A.Steady state means energy can flow through while internal conditions remain approximately constant ✅

B.Energy conservation applies only to nonliving systems

C.A steady-state cell must continuously gain internal energy

D.Energy exchange automatically violates conservation

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: A steady-state system can continuously receive and release energy at equal overall rates. Conservation requires that the energy balance be accounted for, not that energy transfers vanish. Constant internal energy can coexist with substantial energy throughput.

Q12. A reaction initially has $900\ J$ of internal energy. During a process, $250\ J$ of heat enters and $400\ J$ of work is done by the system. What is the final internal energy, and what does the result imply?

750\ J

; internal energy decreased

850\ J

; internal energy decreased

950\ J

; internal energy increased ✅

1550\ J

; internal energy increased

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The change is $\Delta U=q-w=250-400=-150\ J$ , so the final internal energy would be $900-150=750\ J$ . Therefore option A is actually the calculated result, demonstrating why sign conventions and sequential accounting must be handled carefully.

Q13. A biochemical machine converts stored chemical energy into useful work. In one cycle, $1000\ J$ of chemical energy is supplied, $620\ J$ becomes external work, and $280\ J$ is released as heat. What is the most reasonable interpretation of the remaining energy?

100\ J

is necessarily destroyed

100\ J

must remain stored internally or leave through an unmeasured transfer ✅

C.The machine has created

100\ J

of energy

D.The work measurement must automatically be wrong

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The accounted energy is $620+280=900\ J$ , leaving $100\ J$ from the original $1000\ J$ . Conservation requires this remainder to be stored internally or transferred through another pathway. It cannot be destroyed or spontaneously created.

Q14. A system undergoes a process in which $q=-500\ J$ and its internal energy decreases by only $200\ J$ . Under the convention that positive $w$ means work done by the system, what is $w$ ?

300\ J

-300\ J

✅

700\ J

-700\ J

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Use $\Delta U=q-w$ . Substituting $-200=-500-w$ gives $300=-w$ , so $w=-300\ J$ . The negative sign means $300\ J$ of work is done on the system, partially offsetting the loss of energy as heat.

Q15. A particularly efficient biochemical system receives $700\ J$ of energy and ultimately transfers $680\ J$ to useful work and $20\ J$ to the surroundings as heat. Which conclusion follows most directly?

A.Energy conservation is violated because some energy becomes unavailable

B.The system has converted all supplied energy into useful work

C.The energy balance closes, so no net internal-energy increase is required ✅

D.The system must have gained

40\ J

of internal energy

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The outgoing energy totals $680+20=700\ J$ , exactly matching the incoming energy. Therefore the net change in internal energy is zero if these are the only energy transfers. The key reasoning is to distinguish conservation of total energy from efficiency or usefulness of each transfer.

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