Question

In a certain process, $8.00$ kcal of heat is furnished to the system while the system does $6.00 \mathrm{~kJ}$ of work. By how much does the internal energy of the system change during the process? Here $8.00$ kcal is heat-in and $6.00 \mathrm{~kJ}$ is work-out, both of which are positive. Consequently, $\Delta Q=(8000$ cal $)(4.184 \mathrm{~J} / \mathrm{cal})=33.5 \mathrm{~kJ}$ and $\Delta W=6.00 \mathrm{~kJ}$ Therefore, from the First Law $\Delta Q=\Delta U+\Delta W$, $$\Delta U=\Delta Q-\Delta W=33.5 \mathrm{~kJ}-6.00 \mathrm{~kJ}=27.5 \mathrm{~kJ}$$

Name: Problem 1, Chapter 20: Schaum’s Outline of College Physics
Uploaded: 2023-08-01T17:39:30-08:00
Duration: 1 min 31 s
Channel: Supratim Pal

   In a certain process, $8.00$ kcal of heat is furnished to the system while the system does $6.00 \mathrm{~kJ}$ of work. By how much does the internal energy of the system change during the process?
Here $8.00$ kcal is heat-in and $6.00 \mathrm{~kJ}$ is work-out, both of which are positive. Consequently,
$\Delta Q=(8000$ cal $)(4.184 \mathrm{~J} / \mathrm{cal})=33.5 \mathrm{~kJ}$ and $\Delta W=6.00 \mathrm{~kJ}$
Therefore, from the First Law $\Delta Q=\Delta U+\Delta W$,
$$\Delta U=\Delta Q-\Delta W=33.5 \mathrm{~kJ}-6.00 \mathrm{~kJ}=27.5 \mathrm{~kJ}$$

Schaum’s Outline of College Physics

Eugene Hecht 12th Edition

Chapter 20, Problem 1

Instant Answer

Solved by Expert Supratim Pal

08/01/2023

Step 1

We know that 1 calorie is equal to 4.184 joules. Therefore, we can convert 8.00 kcal to joules using the formula: \[\Delta Q = (8.00 \text{ kcal})(1000 \text{ cal/kcal})(4.184 \text{ J/cal}) = 33.5 \text{ kJ}\] Show more…

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In a certain process, $8.00$ kcal of heat is furnished to the system while the system does $6.00 \mathrm{~kJ}$ of work. By how much does the internal energy of the system change during the process? Here $8.00$ kcal is heat-in and $6.00 \mathrm{~kJ}$ is work-out, both of which are positive. Consequently, $\Delta Q=(8000$ cal $)(4.184 \mathrm{~J} / \mathrm{cal})=33.5 \mathrm{~kJ}$ and $\Delta W=6.00 \mathrm{~kJ}$ Therefore, from the First Law $\Delta Q=\Delta U+\Delta W$, $$\Delta U=\Delta Q-\Delta W=33.5 \mathrm{~kJ}-6.00 \mathrm{~kJ}=27.5 \mathrm{~kJ}$$

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Key Concepts

First Law of Thermodynamics

This principle states that energy is conserved in a system, establishing that the change in internal energy equals the heat added to the system minus the work done by the system. It is fundamental for determining how energy exchanges affect the state of a system.

Work and Heat Sign Conventions

Understanding the sign conventions for work and heat in thermodynamics is essential. Typically, heat added to a system is taken as positive, and work done by the system is considered positive unless specified otherwise, influencing how energy changes are calculated.

Energy Unit Conversions

Consistent unit usage is crucial in thermodynamics. Converting energy units, such as from calories to joules or kilojoules, ensures accurate computations when applying the First Law, as it prevents errors that can arise from unit mismatches.

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