Question

Suppose 200 Jof work is done on a system and 70.0 calis extracted from the system as heat. In the sense of the first law of thermodynamics, what are the values (including algebraic signs) of (a) W, (b) Q, and (c)$\mathbf{∆}{\mathbf{E}}_{\mathbf{int}}$?

Short answer

1. The value of W is$-200\mathrm{J}$
2. The value of Q is$-293\mathrm{J}$
3. The value of$∆{\mathrm{E}}_{\mathrm{int}}$ is$-93\mathrm{J}$

Step-by-step solution

The given data

1. Work done on the system is$200\mathrm{J}$
2. Heat extracted from the system is$-70\mathrm{cal}$

Understanding the concept of first law of thermodynamics

According to the first law of thermodynamics, energy can only be transformed, not created or destroyed. Using the first law of thermodynamics, we can calculate the above quantities.

Formula:

The equation to the heat change in the first law of thermodynamics,

$\mathrm{dQ}=\mathrm{dW}+{\mathrm{dE}}_{\mathrm{int}}$ …(i)

(a) Calculation of the work done

W is positive if work is done by the system, and negative if work is done on the system. Since the work is done on the system, it is taken as negative.

So,

$\mathrm{W}=-200\mathrm{J}$

Hence, the value of W is$-200\mathrm{J}$

(b) Calculation of heat Q

Q is positive if heat is added to the system and negative if heat is removed. Since the heat is extracted from the system, it is taken as negative.

Therefore,

$\begin{array}{rcl}\mathrm{Q}& =& -70\mathrm{cal}\\ & =& -70\times 4.18\mathrm{J}\left(\because 1\mathrm{cal}=4.18\mathrm{J}\right)\\ & =& -292.6\mathrm{J}\\ & \approx & -293\mathrm{J}\end{array}$

Hence, the value of the heat, Q isdata-custom-editor="chemistry" $-293\mathrm{J}$

(c) Calculation of the internal energy ∆Eint 

From first law of thermodynamics, that is equation (i), we get the internal energy of the system as:

$\begin{array}{rcl}∆{\mathrm{E}}_{\mathrm{int}}& =& \mathrm{Q}-\mathrm{W}\\ & =& -293\mathrm{J}-\left(-200\mathrm{J}\right)\\ & =& -93\mathrm{J}\end{array}$

Hence, the value of the internal energy is$-93\mathrm{J}$.