Question

A system is provided with the \[50\] Joules of heat and the work done on the system is \[10\] Joules. What is the change in the internal energy of the system in Joules?
\[(i){\text{ 60}}\]
\[(ii){\text{ 40}}\]
\[(iii){\text{ 50}}\]
\[(iv){\text{ 10}}\]

Answer 1

Hint: Here the heat is provided to the system and work is done on the system. We will use the law of thermodynamics to find the change in internal energy. According to the law of thermodynamics, the change in internal energy of a system is the sum of heat transferred and work done.

Complete answer:
The internal energy refers to all the energy of the system. We also know from the first law of thermodynamics that energy can neither be created nor be destroyed. It can change from one form to another form. Thus the relation between the change in internal energy, work done and heat of the system can be depicted as,
\[\Delta U{\text{ = Q + W}}\]
According to it, the change in internal energy is equal to the sum of heat transferred and work done.
According to the question, work is done on the system, therefore the value of work done is positive since energy is being supplied to the system. Also when we provide heat to the system it is taken as a positive value. Hence we concluded that \[Q{\text{ = + 50 J}}\] and \[{\text{W = + 10 J}}\]. On substituting the given values in the above equation we get,
\[\Delta U{\text{ = Q + W}}\]
\[\Delta U{\text{ = }}\left( {{\text{50 + 10}}} \right){\text{ J}}\]
\[\Delta U{\text{ = 60 J}}\]
Hence the change in internal energy is \[60\]Joules. Hence the correct option is \[(i){\text{ 60}}\].

Note:
When the heat is provided to the system then it is taken as positive but when heat is absorbed from the system then it is negative. Similarly when work is done on a system then it is taken as positive but when work is done by system then it is taken as negative. Since it is an energy its unit is Joule.