Question

In a thermodynamic process, If the amount of work done on the gas by its surrounding is
$320\;{\rm{J}}$ and the internal energy is increased by $560\;{\rm{J}}$. Calculate how much heat is transferred between the gas and its surrounding.
A. $240\;{\rm{J}}$ absorbed
B. $240\;{\rm{J}}$ dissipated
C. $880\;{\rm{J}}$ absorbed
D. $880\;{\rm{J}}$ dissipated

Answer 1

The above problem can be resolved using the fundamentals of thermodynamics and heat transfer. The thermodynamics is that branch of thermal analysis, where the heat interaction form system takes place. There are three basic thermodynamics laws, which provide the vast knowledge in obtaining the significant relationship between energy transfer, the internal energy, and the work obtained due to the transfer of energy.

Complete step by step solution:
Apply the first law of thermodynamics as,
$Q = \Delta U + W$
Here, Q is the thermal energy transfer, U is the internal energy and W is the work interaction.
As per the given condition, the work is done on the system so, the sign convention for the work done will be negative. And as there is the increase in the internal energy of the system, then the sign convention for the internal energy is positive.
Substitute the value as,

Q = \Delta U + W\\\ \Rightarrow Q = 560\;{\rm{J}} + \left( { - 320\;{\rm{J}}} \right)\\\ \Rightarrow Q = 240\;{\rm{J}} \end{array}$$ As, the sign convention for the heat transfer is positive, then the heat will be absorbed by the system. **Therefore, 240 joules of heat is transferred between the gas and its surrounding and option (A) is correct.** **Note:** Try to understand the concept and applications of the laws of thermodynamics. The first law of thermodynamics tells us that, when the system is supposed to contain some energy, then some part of that energy is convertible to the work. The left out part of the energy is being utilized to increase the internal energy of the system.