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
654.6k+ views
Hint: 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,
\[\begin{array}{l}
Q = \Delta U + W\\
\Rightarrow Q = 560\;{\rm{J}} + \left( { - 320\;{\rm{J}}} \right)\\
\Rightarrow Q = 240\;{\rm{J}}
\end{array}\]
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.
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,
\[\begin{array}{l}
Q = \Delta U + W\\
\Rightarrow Q = 560\;{\rm{J}} + \left( { - 320\;{\rm{J}}} \right)\\
\Rightarrow Q = 240\;{\rm{J}}
\end{array}\]
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.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

