Question

An electric heater supplies heat to a system at a rate of 100 watts. If a system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing?

Answer 1

Hint: Apply the first law of thermodynamics and find the internal energy in terms of work done and heat supplied. Using the given Value of heat and work done and find out the change in internal energy of the system. Take care of the signs.

Step by step solution:
We are given the heat supplied to the system is: 100 W
As the heat is supplied to the system we have to take $\Delta$ Q as the positive value.
If the heat was released by the system then we have to take $\Delta$ Q as the negative value .
So, in this case, we get the change in the heat energy $\Delta$ Q = 100 W
Similarly, work done by the system is given as 75 $\dfrac{J}{s}$
We know if the system has done the work so we need to take the positive value and if the work has been done by others on the system then we have to take this sign as negative
As per the law of thermodynamics, we know that :
change in the heat energy is equal to the sum of the change in internal energy and work done by the system
So we can write $\Delta Q = \Delta U + \Delta W$
Substituting the values we get the change in the internal energy as:
$\Delta U = \Delta Q - \Delta W$
$\Delta U$ = 100 – 75 = 25 $\dfrac{J}{s}$
thus using the first law of thermodynamics we have calculated the rate of change of internal energy as 25 joules per second

Additional Information:
The Law of conservation of energy in thermodynamics is called the first law of thermodynamics which states that energy neither can be created nor destroyed and energy can only be transferred or changed from one form to another form.
For example we see that the laptop which we use is storing the energy and then using it and gives us the energy in another form which is the light energy from the screen and the other forms of the output of the laptop is the manifestation of the energy we supplied.
This law says that if the heat flows into the system the internal energy of the system increases and we take the sign of the q and w as positive and if the heat is flowing out of the system we say that the system is doing work and the sign of the q and w will be negative.

Note: In this kind of problem we need to take care of signs to be used for the change in internal energy work done and change in heat energy otherwise the answer will go wrong. Try to differentiate between heat supplied and the heat gained by the system and work done and the work done upon the system.