Question

Heat of 30 kcal is supplied to a system and 4200 J of external work is done on the system so that its volume decreases at constant pressure. What is the change in its internal energy? (J = 4200 J/kcal)
A.302 $ \times {10^5}$
B.302 $ \times {10^5}$
C.302$ \times {10^5}$
D.302$ \times {10^5}$

Answer 1

Hint:The heat supplied to the system is considered to be positive. Since the volume is decreasing, the work done on the system is negative.

Complete step by step solution:
The heat supplied to the system is given to us in kcal which is not the SI unit of heat. So we first need to convert heat into joules. For that, we multiply the heat given in cal with 4200. The heat supplied to a system is denoted as $\Delta $Q.
$\Delta $Q = 30 kcal $ \times $ 4200 J/kcal = 126000 J
The work done on the system is such that the volume of the system is decreasing. So the work done on the system is negative. Work done is denoted by $\Delta $W.
$\Delta $W = - 4200 J
We now apply the first law of thermodynamics to find out the change in internal energy of the system which is denoted as $\Delta $U. The first law of thermodynamics states that the heat transfer into a system is equal to the sum of the change in internal energy of the system and work done by the system.
$\Delta Q = \Delta U + \Delta W$ …equation (1)
On substituting the values of heat transferred and work done in equation (1), we obtain,
126000 = $\Delta $U + (- 4200)
Upon simplifying, we obtain the value of the internal energy, which is,
$\Delta $U = 126000 – (- 4200) = 126000 + 4200 = 130200 J = 1.302$ \times {10^5}$ J
Hence, the change in the internal energy of the system is 1.302$ \times {10^5}$ J.

Therefore, the correct answer for this question is option A.

Note: The sign convention for thermodynamics to be used is:
-Q is positive if heat is supplied to the system
-Q is negative if heat is taken away from the system
-W is positive if work is done by the system
-W is negative if work is done on the system