Question

Find the change in the internal energy when $15gm$of air is heated from ${0^ \circ }C$to${5^ \circ }C$. The specific heat of air at constant volume is $0.2{\text{ cal(g}}{{\text{m}}^ \circ }{\text{C}}{{\text{)}}^{ - 1}}$
A. $75{\text{ cal}}$
B. ${\text{30 cal}}$
C. $15{\text{ cal}}$
D. $105{\text{ cal}}$

Answer 1

Hint:The specific heat at constant volume is the amount that is needed to raise the temperature of unit mass of a gas by one degree at constant volume. Specific heat capacity at constant pressure is greater than the specific heat capacity at constant volume because at constant pressure, some of the energy goes into raising the temperature and some of the energy goes into doing work by expanding the ideal gas.

Formula used:
$\Delta Q = mc\Delta T$
Where $m = {\text{mass of substance}}$, $\Delta Q = {\text{change in heat}}$, $c = {\text{specific heat}}$ and $\Delta T = {\text{change in temperature}}$

Complete step by step answer:
Whenever heat energy is added to a substance, the temperature will change and the relationship between heat energy and temperature is different for each material and the specific heat is used to describe how they are related. It is given by the formula:
$\Delta Q = mc\Delta T$
Now,
$\Delta Q = mc\Delta T \\
\Rightarrow \Delta Q= {\text{Mass}} \times {\text{specific heat}} \times {\text{temperature change}}$
$\Rightarrow \Delta Q = 15 \times 0.2 \times 5\,cal$
$\therefore \Delta Q = 15\,cal$
For an isochoric process, work done is equal to zero.
By applying first law, the change in internal energy$ = \Delta Q = 15\,cal$

Therefore, option C is the correct answer.

Note:The change in the internal energy of a system is the sum of the heat transferred to the system and the work done. When the volume of a given system is constant, the change in its internal energy can be calculated by substituting the ideal gas law into the equation for change in internal energy.