Chapter 6.1, Problem 6.1BFP

Interpretation Introduction

Interpretation:

$\Delta E$ of a system when surroundings absorbs $26.0\text{ kcal}$ of heat and work done on the system is $15.0\text{ Btu}$ is to be calculated.

Concept introduction:

The first law of thermodynamics gives the relation between the work, heat and internal energy. According to this law “the total energy of the system plus surrounding remain constant”. The expression to first law of thermodynamics is:

$\Delta E=q+w$

Here,

$\Delta E$ is the internal energy of the system.

$q$ is the heat released or absorbed.

$w$ is the work done by the system or on the system.

The work done by the system and heat released by the system is negative and the work done on the system and heat taken by the system is positive.

The expression to convert $\text{kcal}$ into $\text{kJ}$ is:

$\text{1}\text{kcal}=4.184\text{kJ}$

The expression to convert $\text{Btu}$ into $\text{kJ}$ is:

$\text{1}\text{Btu}=1.055\text{kJ}$