100g of water is heated from `30^@C to 50^@C`. Ignoring the slight expansion of the water, the change in its internal energy is (specific heat of water is `4184J//kg//K`):

Options

A. `8.4kJ`
B. `84kJ`
C. `2.1kJ`
D. `4.2kJ`

Explanation

To find the change in internal energy of 100g of water heated from 30°C to 50°C, we can follow these steps:

### Step 1: Identify the mass of water
The mass of water (M) is given as 100 grams. We need to convert this to kilograms because the specific heat capacity is given in J/(kg·K).

\[ M = 100 \, \text{g} = \frac{100}{1000} \, \text{kg} = 0.1 \, \text{kg} \]

**Hint:** Always convert grams to kilograms when using specific heat capacity in J/(kg·K).

### Step 2: Determine the change in temperature
The initial temperature (T1) is 30°C and the final temperature (T2) is 50°C. The change in temperature (ΔT) can be calculated as:

\[ \Delta T = T2 - T1 = 50°C - 30°C = 20°C \]

**Hint:** The change in temperature is simply the difference between the final and initial temperatures.

### Step 3: Use the formula for heat transfer
The heat gained by the water (Q) can be calculated using the formula:

\[ Q = M \cdot C \cdot \Delta T \]

Where:
- \(C\) is the specific heat capacity of water, which is given as \(4184 \, \text{J/(kg·K)}\).

Substituting the values we have:

\[ Q = 0.1 \, \text{kg} \cdot 4184 \, \text{J/(kg·K)} \cdot 20 \, \text{K} \]

### Step 4: Calculate the heat transfer
Now, we can calculate \(Q\):

\[ Q = 0.1 \cdot 4184 \cdot 20 = 8368 \, \text{J} \]

### Step 5: Relate heat transfer to change in internal energy
According to the first law of thermodynamics, the change in internal energy (ΔU) is given by:

\[ \Delta U = Q - W \]

Since we are ignoring the work done (W = 0, due to no volume change), we have:

\[ \Delta U = Q \]

Thus,

\[ \Delta U = 8368 \, \text{J} \]

### Step 6: Convert to kilojoules
To express the change in internal energy in kilojoules:

\[ \Delta U = \frac{8368}{1000} = 8.368 \, \text{kJ} \]

### Final Answer
The change in internal energy is approximately:

\[ \Delta U \approx 8.4 \, \text{kJ} \]

### Summary
The change in internal energy of the water when heated from 30°C to 50°C is approximately 8.4 kJ.

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