Question

Amount of 62 J of work is done on the system and 128 J of heat is transferred to
the surroundings. What is the internal energy change for the above process?

a) 66J b)190J c) -190J d) -66J

Answer 1

Answer:

(d)-66J

Explanation:

Work done on the system=62J

or w=62J

Heat given out by the system=128J

or q= -128J

Now u=q+w=-128+62=-66J

Thus internal energy of the system decreases by -66J

Answer 2

Answer:

The change in internal energy for the given process is option $(\ d\ )$, $-\ 66$ J·

Explanation:

Given that,

The amount of work done on the system, W $=$ $+\ 62$ J

The amount of heat transferred to  the surrounding, q  $=$ $-128$ J

Since,

The work is done on the system we took the value as positive and the heat is transferred to the surroundings the value took it as negative·

To find out the change in internal energy,

From the first law of thermodynamics we have

   $\Delta U\ =\ q\ +\ W$   $-------(1)$

Here, the value of q is negative and the value of W is positive· So the equation becomes,

   $\Delta U\ =\ (\ -q\ )\ +\ W$    $-------(2)$

On substituting the values in equation No $2$, we get

⇒ $\Delta U\ =\ (\ -\ 128\ )\ +\ 62$

⇒ $\Delta U\ =\ -\ 66$ J

Therefore,

The change in internal energy for the given process is $-\ 66$ J·