Phase transitions are ubiquitous in nature. We are all familiar with the different phase of water (vapour, liquid and ice) and with the change from one to another, the change of phase are called phase transitions. There are six ways a substance can change between these three phase, melting, freezing, evaporating, condensing sublimation and decomposition.
At `1 atm` pressure vaporisation of `1` mole of water from liquid `(75^(@)C)` to vapour `(120^(@)C)`.
`C_(v)(H_(2)O,l)=75 J "mole"^(-1)K^(-1), C_(p)(H_(2)O,g)=33.3J"mole"^(-1)K^(_1)`
`Delta H_(vap)` at `100^(@)C=40.7KJ//"mole"`
Calculate change in internal energy when
Water liquid at `75^(@)C` to `100^(@)C` ?

Options

A. `1875 J`
B. `13125 J`
C. `-1875 J`
D. `-13125 J`

Answer

To calculate the change in internal energy when water changes from liquid at \(75^{\circ}C\) to liquid at \(100^{\circ}C\), we will use the formula for change in internal energy (\(\Delta U\)):

\[ \Delta U = n C_v \Delta T \]

### Step-by-Step Solution:

**Step 1: Identify the variables.**
- Number of moles (\(n\)) = 1 mole (given in the question)
- Heat capacity at constant volume (\(C_v\)) = 75 J mole\(^{-1}\) K\(^{-1}\) (given in the question)
- Initial temperature (\(T_i\)) = \(75^{\circ}C\)
- Final temperature (\(T_f\)) = \(100^{\circ}C\)

**Step 2: Calculate the change in temperature (\(\Delta T\)).**
\[ \Delta T = T_f - T_i = 100^{\circ}C - 75^{\circ}C = 25^{\circ}C \]
Since \(1^{\circ}C\) change is equivalent to \(1 K\) change, we can also express this as:
\[ \Delta T = 25 K \]

**Step 3: Substitute the values into the formula for \(\Delta U\).**
\[ \Delta U = n C_v \Delta T = 1 \, \text{mole} \times 75 \, \text{J mole}^{-1} \text{K}^{-1} \times 25 \, \text{K} \]

**Step 4: Perform the calculation.**
\[ \Delta U = 1 \times 75 \times 25 = 1875 \, \text{J} \]

### Final Answer:
The change in internal energy (\(\Delta U\)) when water changes from liquid at \(75^{\circ}C\) to liquid at \(100^{\circ}C\) is **1875 J**.

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