Calculate the change in internal energy of the following system: a 100.0-g bar of gold is heated from 25 ∘C to 50 ∘C during whic

Question

Damm

1 month ago

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Calculate the change in internal energy of the following system: a 100.0-g bar of gold is heated from 25 ∘C to 50 ∘C during whic

h it absorbs 322 J of heat. Assume the volume of the gold bar remains constant.

Chemistry

2 answers:

castortr0y [3K] · Answer 1 · 2026-04-04T12:53:00+03:00

According to the Law, the variation in internal energy (U) of the system is equal to the total of the heat added to the system (q) plus the work performed ON the system (W)
ΔU = q + W 

In response to the first question, 0.653 kJ of heat energy is extracted from the system (balloon) while 386 J of work is applied to the balloon, leading to 
ΔU = -653J + 386J 
=-267J 
Thus, the internal energy reduces by 267 J 

For the second question, 322 J of heat is supplied to the system (gold bar) while no work is undertaken on the gold bar, marking this as an isochoric/isovolumetric process, thus 
ΔU = 322J + 0 
=322J 
Hence, internal energy rises by 322 J

Tems11 [2.7K] · Answer 2 · 2026-04-04T12:53:53+03:00

The internal energy change of a gold bar is $\boxed{{\text{322 J}}}$

Additional clarification:

Thermodynamics:

It refers to a chemistry segment related to heat temperature and its relationship with work, termed thermodynamics, which finds extensive applications in pyrometallurgy, gas phase reactions, solid-gas interactions, ATP production, among other areas.

First law of Thermodynamics:

It hinges on the conservation of energy principle, asserting that the total energy within a system remains unchanged. According to this principle, the variation in internal energy is computed based on the total work performed on the system and the heat added. Its mathematical representation is as follows:

$\Delta {\text{U}}={\text{q}}+{\text{W}}$ …… (1)

Here,

$\Delta{\text{U}}$ signifies the change in internal energy of the system.

q is the heat supplied to the system.

W refers to the work done on the system.

The equation for calculating work performed can be expressed as:

${\text{W}}={\text{P}}\Delta{\text{V}}$ …… (2)

Where,

W indicates the work undertaken.

P represents the pressure on the system.

$\Delta{\text{V}}$ signifies the change in volume of the system.

Given that the gold bar's volume remains constant, it indicates $\Delta{\text{V}}$ to be zero.

Substituting 0 for $\Delta{\text{V}}$ in equation (2) allows for the calculation of work done on the gold bar.

$\begin{aligned}{\text{W}}&={\text{P}}\left( 0 \right)\\&=0\\\end{aligned}$

The value of q stands at 322 J.

The value of W is 0.

Placing these values in equation (1) enables the calculation of the change in internal energy of the gold bar.

$\begin{gathered}\Delta{\text{U}}={\text{322 J}}+0\\={\text{322 J}}\\\end{gathered}$

Consequently, the change in internal energy of the gold bar amounts to 322 J.

Explore further:

1. What is the final temperature of copper?

2. Calculate the specific heat of gold:

Response details:

Grade: Senior School

Subject: Chemistry

Chapter: Thermodynamics

Keywords: internal energy, heat, work, q, W, P, 322 J, 0, thermodynamics, first law of thermodynamics, law of conservation of energy, total energy, isolated system, volume, constant.