What is the change in internal energy (in \(J\) ) of a system that releases \(675 \mathrm{~J}\) of thermal energy to its surroundings and has \(530 \mathrm{cal}\) of work done on it?

Short Answer

Expert verified
The change in internal energy is 1542.52 J.

Step by step solution

01

- Understand the Given Data

Identify the quantities provided in the problem:- Thermal energy released by the system: \[ q = -675 \text{ J} \]- Work done on the system (in calories): \[ w = 530 \text{ cal} \]
02

- Convert Work from Calories to Joules

Convert the work done on the system into joules using the conversion factor: \[ 1 \text{ cal} = 4.184 \text{ J} \]So, \[ w = 530 \text{ cal} \times 4.184 \text{ J/cal} = 2217.52 \text{ J} \]
03

- Use the First Law of Thermodynamics

According to the first law of thermodynamics, the change in internal energy \( \text{ΔU} \) can be calculated using the formula:\[ \text{ΔU} = q + w \]
04

- Calculate the Change in Internal Energy

Now substitute the values of \( q \) and \( w \) into the formula:\[ \text{ΔU} = -675 \text{ J} + 2217.52 \text{ J} \]\[ \text{ΔU} = 1542.52 \text{ J} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

thermal energy
Thermal energy refers to the internal energy present in a system due to the random motions of its particles. This energy is related to the temperature of the system and is a fundamental concept in thermodynamics.
In the context of our exercise, thermal energy is released by the system, which means the system is losing heat to its surroundings. Let's denote this by the symbol \( q \).
  • If \( q \) is positive ( \( q > 0 \)), the system absorbs heat.
  • If \( q \) is negative ( \( q < 0 \)), the system releases heat.
In our exercise, \( q = -675\text{ J} \) because the system is releasing 675 joules of thermal energy to its surroundings.
first law of thermodynamics
The First Law of Thermodynamics is a fundamental principle, often known as the law of energy conservation. It states that the total energy of an isolated system is constant. Energy can neither be created nor destroyed, but it can change forms.
Mathematically, it is represented as:
\[ \Delta U = q + w \] where:
  • \( \Delta U \) is the change in internal energy.
  • \( q \) is the heat added to the system (thermal energy).
  • \( w \) is the work done on the system.
In our context, the work ( \( w \)) done on the system is given in calories. Work done on the system increases its internal energy (hence positive). Using the First Law, we combine the thermal energy change and work done to find the overall change in internal energy.
calories to joules conversion
In many thermodynamic problems, energy may be given in different units. It’s essential to convert these units to perform accurate calculations.
The conversion factor between calories and joules is:
\[ 1 \text{ cal} = 4.184 \text{ J} \] For our problem, the work done on the system is given as 530 calories.
We need to convert this to joules to use in our formula:
  • \[ w = 530 \text{ cal} \times 4.184 \text{ J/cal} = 2217.52 \text{ J} \]
By converting the calories to joules, we ensure all our values are in the same units, making our calculations straightforward. Therefore, substituting into the First Law equation, we get:
\[ \Delta U = q + w \] \[ = -675 \text{ J} + 2217.52 \text{ J} \]
Resulting in \[ \Delta U = 1542.52 \text{ J} \].

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