Question

Find the external work done by the system in kcal, when 20 kcal of heat is supplied to the system the increase in internal energy is 8400 J (${\text{J = 4200J/kcal}}$)
(A) 16 kcal
(B) 18 kcal
(C) 20 kcal
(D) 19 kcal

Answer 1

Hint : When work is done by a system, it reduces the internal energy of that system. Application of heat works to raise the internal energy of the system. By substituting the values in the question into the formula, $W = \Delta Q - \Delta U$ we will get the answer.

Formula used: In this solution we will be using the following formula;
$\Rightarrow W = \Delta Q - \Delta U$ where, $W$ is the work done by the system, $\Delta Q$ is the heat applied to the system, and is the change in internal energy of the system.

Complete step by step answer
When heat is applied to a system, the temperature of the system tends to increase, hence so does the internal energy. This is because the internal energy can be said to be a measure of the temperature of a system. But when a system is allowed to do work (on the surrounding), the internal energy reduces since it must have used some of the internal energy. The statement which guides such process is given by the first law of thermodynamics, and can be mathematically given, in one form, as
$\Rightarrow \Delta U = \Delta Q - W$which implies that the increase in internal energy of a system is equal to the increase in thermal energy (heat applied) minus the work done by the system.
It can be rearranged to be
$\Rightarrow W = \Delta Q - \Delta U$
Hence, we can insert known values. Before that we must convert the internal energy to kcal
Hence, $8400{\text{J = 8400J}} \times \dfrac{{\text{1}}}{{4200}}\dfrac{{kcal}}{{\text{J}}} = 2kcal$
Now, by substitution, we have
$\Rightarrow W = 20 - 2 = 18kcal$
Hence, the correct option is B.

Note
For clarity, it isn’t uncommon to see the first law of thermodynamics written as
$\Rightarrow \Delta U = \Delta Q + W$. The resolution of this seeming contradiction lies in the definition of $W$.
When $W$ is defined as the work done by the system i.e. it is positive when work is done by the system then we write it as it is written in the step by step solution. But when $W$ is defined as the work done on the system i.e. positive when work is done on the system, it is written as in the note, i.e. $\Delta U = \Delta Q + W$