This question involves calculating the change in internal energy when water is heated. Since water is being heated at constant volume (ignoring expansion), the heat added goes entirely into increasing its internal energy, with no work done.

Step 1: Understand the Concept
For a process at constant volume, the first law of thermodynamics states:
$\Delta U=Q$
where $\Delta U$ is the change in internal energy and $Q$ is the heat added. There is no work done ($W=0$) because the volume is constant.

Step 2: Calculate the Heat Added (Q)
The heat required to change the temperature of a substance is given by:
$Q=mc\Delta T$
where:
$m$ = mass
$c$ = specific heat capacity
$\Delta T$ = change in temperature

Step 3: Plug in the Values
Given:
Mass, $m=100\mathrm{g}=0.1\mathrm{kg}$
Specific heat, $c=4184\mathrm{J}/\mathrm{kg}/\mathrm{K}$
Temperature change, $\Delta T=50°\mathrm{C}-30°\mathrm{C}=20\mathrm{K}$ (since the size of a degree Celsius is the same as a Kelvin)

Now calculate Q:
$Q=\left(0.1\right)\times \left(4184\right)\times \left(20\right)$
First, multiply 0.1 and 20: $0.1\times 20=2$
Then, multiply by 4184: $2\times 4184=8368\mathrm{J}$

Step 4: Convert to Kilojoules
$8368\mathrm{J}=8.368\mathrm{kJ}\approx 8.4\mathrm{kJ}$
This value is approximately 8.4 kJ.

Final Answer: The change in internal energy is approximately 8.4 kJ.

Related Topics and Formulae

First Law of Thermodynamics:
$\Delta U=Q-W$
where $Q$ is the heat added to the system and $W$ is the work done by the system.

Heat Transfer at Constant Volume:
For constant volume processes ($\Delta V=0$), no work is done ($W=0$). Therefore, $\Delta U=Q=mc\Delta T$.

Specific Heat Capacity:
The specific heat capacity ($c$) is the amount of heat required to raise the temperature of 1 kg of a substance by 1 K.