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. $268799.2$
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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

Step 1: Calculating heat supplied to water.
\[ Q = mc\Delta T \] \[ Q = 4 \times 4200 \times (20-4) \] \[ Q = 268800\,\text{J} \] Step 2: Calculating work done due to expansion.
Volume at $4^\circ$C:
\[ V_1 = \dfrac{4}{1000} = 0.004\,\text{m}^3 \] Volume at $20^\circ$C:
\[ V_2 = \dfrac{4}{998} = 0.004008\,\text{m}^3 \] Step 3: Change in volume.
\[ \Delta V = V_2 - V_1 = 8 \times 10^{-6}\,\text{m}^3 \] Step 4: Work done at constant pressure.
\[ W = P\Delta V = 10^5 \times 8 \times 10^{-6} = 0.8\,\text{J} \] Step 5: Increase in internal energy.
\[ \Delta U = Q - W \] \[ \Delta U = 268800 - 0.8 = 268799.2\,\text{J} \] Step 6: Final conclusion.
The increase in internal energy is $268799.2\,\text{J}$.