Question

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

Answer 1

The Law states that the change in internal energy (U) of the system is equal to the sum of the heat supplied to the system (q) and the work done ON the system (W) 
ΔU = q + W 

For the first question, 0.653kJ of heat energy is removed from the system (balloon) while 386J of work is done ON the balloon, thus 
ΔU = -653J + 386J 
=-267J 
Thus internal energy decrease by 267J 

For the second question, 322J of heat energy is added to the system (gold bar) while no work is done on the gold bar, this is an isochoric/isovolumetric process, thus 
ΔU = 322J + 0 
=322J 
Thus internal energy increase by 322J

Answer 2

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

Further explanation:

Thermodynamics:

It is branch of chemistry that is related to heat temperature and connection with work is known as thermodynamics it has wide uses in pyrometallurgy, gas phase reaction gas solid reaction and ATP generation and many more.

First law of Thermodynamics:

It is based on law of energy conservation that states total energy of system remains constant. According to this law, internal energy change is determined by sum of work done on the system and heat supplied. Its mathematical expression is as follows:

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

Here,

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

q is the heat supplied to the system.

W is the work done on the system.

The formula to calculate the work done is as follows:

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

Here,

W is the work done.

P is the pressure on the system.

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

It is given that the volume of the gold bar remains constant. This implies $\Delta{\text{V}}$ to be zero.

Substitute 0 for $\Delta{\text{V}}$ in equation (2) to calculate the work done on the gold bar.

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

The value of q is 322 J.

The value of W is 0.

Substitute these values in equation (1) to calculate the change in the internal energy of the gold bar.

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

So the change in internal energy of gold bar is 322 J.

Learn more:

1. What is the final temperature of copper? brainly.com/question/7210400

2. Calculate the specific heat of gold: brainly.com/question/3032746

Answer 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.