Question

Given, 200gm of water is heated from ${{20}^{0}}C$ to ${{40}^{0}}C$. Ignoring the slight expansion of water, change in internal energy is equal to (in kJ)

Answer 1

Hint: The first law of thermodynamics can be used for the above question. As the expansion of the water is told to be ignored, work done by the system is zero meaning three changes in internal energy being equal to the heat released.

Formulas used:
$\Delta U=Q$

Complete answer:
Let us first find the work done by the system.
The work done by the system is zero as the expansion of water is neglected or ignored,
Therefore, applying this in the first law of thermodynamics,
Internal energy= heat released
Now, the heat released by the system will be equal to,
$\begin{align}
  & Q=ms\Delta T \\
 & \Rightarrow Q=\dfrac{200}{1000}\times 4184\times 20J \\
 & \Rightarrow Q=16736J \\
 & \Rightarrow Q=16.7kJ \\
\end{align}$

Therefore, the correct option is option b.

Additional information:
The first law of thermodynamics is a version of the law of conservation of energy adapted for thermodynamic processes distinguishing 2 kinds of transfer of energy. The energy is distinguished in the form of heat and thermodynamics work, these 2 forms of energy are related to a function of the body's state called internal energy. The law of conservation of energy states that the total energy of an isolated system is constant, energy can be transformed from one form to another, but cannot be created nor destroyed. Sign conventions play an important role in the laws of thermodynamics. In a non-cyclic process, the change in internal energy of a system is equal to net energy added as heat to the system minus the network done by the system.

Note:
In physics, compared to chemistry, the sign conventions are really different. The net work done by the system is taken positively in chemistry and negatively when solving in physics. Therefore, one needs to take care of the signs while solving the work done by the system or for the system in both cases.