Question

Density of water at $4^\circ$C and $20^\circ$C are $1000\,\text{kg/m}^3$ and $998\,\text{kg/m}^3$ respectively. The increase in internal energy of $4\,\text{kg}$ of water when it is heated from $4^\circ$C to $20^\circ$C is ___ J.
(Specific heat capacity of water $= 4.2\,\text{J g}^{-1}\text{K}^{-1}$ and atmospheric pressure $=10^5\,\text{Pa}$)

• A. $315826.2$
• B. $258700.8$
• C. $234699.2$
• D. $268799.2$

Hint:

At constant pressure, $\Delta U = Q - P\Delta V$. For liquids, work done is usually very small.

Answer 1

The Correct Option is D

To calculate the increase in internal energy of water when it is heated from \(4^\circ\text{C}\) to \(20^\circ\text{C}\), we will use the formula:

\(Q = mc\Delta T\) 

where:

• \(m\) = mass of the water = 4 kg = 4000 g
• \(c\) = specific heat capacity of water = 4.2 J/g K
• \(\Delta T\) = change in temperature = \(20^\circ\text{C} - 4^\circ\text{C} = 16\text{ C}\)

Plugging these values into the formula, we get:

\(Q = 4000 \times 4.2 \times 16\)

Calculating further:

\(Q = 4000 \times 67.2 = 268800 \text{ J}\)

Thus, the increase in internal energy of 4 kg of water when heated from \(4^\circ\text{C}\) to \(20^\circ\text{C}\) is approximately 268800 J. The closest option provided is \(268799.2 \text{ J}\).

This matches the answer $268799.2. Therefore, the correct option is:

$268799.2$