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

Explanation

To find the change in internal energy of 100g of water heated from 30°C to 50°C, we can use the formula for heat transfer, which is related to the change in internal energy when no work is done.

### Step-by-Step Solution:

1. **Identify the given values:**
- Mass of water (m) = 100 g = 0.1 kg (since 1 kg = 1000 g)
- Specific heat of water (s) = 4184 J/(kg·K)
- Initial temperature (T1) = 30°C
- Final temperature (T2) = 50°C

2. **Calculate the change in temperature (ΔT):**
\[ \Delta T = T2 - T1 = 50°C - 30°C = 20°C \]
Since the specific heat is given in terms of Kelvin, we can use the same value for Celsius as the change in temperature is the same in both scales.

3. **Use the formula for heat transfer (Q):**
\[ Q = m \cdot s \cdot \Delta T \]
Substituting the known values:
\[ Q = 0.1 \, \text{kg} \cdot 4184 \, \text{J/(kg·K)} \cdot 20 \, \text{K} \]

4. **Calculate Q:**
\[ Q = 0.1 \cdot 4184 \cdot 20 = 8368 \, \text{J} \]

5. **Relate heat transfer to change in internal energy (ΔU):**
Since there is no work done (as stated in the problem), we have:
\[ \Delta U = Q \]
Thus,
\[ \Delta U = 8368 \, \text{J} \]

6. **Final Result:**
The change in internal energy (ΔU) is:
\[ \Delta U = 8368 \, \text{J} \]