Related Formula
Δ Q = mcΔ T$$\Delta Q = mc\Delta T$$
W = P Δ V$W = P \Delta V$
Δ U = Δ Q - W$$\Delta U = \Delta Q - W$$
Core Logic
First, calculate the heat supplied Δ Q$\Delta Q$:
Q = mSΔ T = 4 × 4200 × 16 = 268800 J$$Q = mS\Delta T = 4 \times 4200 \times 16 = 268800 \text{ J}$$
Next, calculate the work done W$W$ against atmospheric pressure due to volume change:
W = P Δ V$W = P \Delta V$
Δ V = ( (m)/(ρf) - (m)/(ρᵢ) ) = 4 [ (1)/(998) - (1)/(1000) ]$$\Delta V = \left( \frac{m}{\rho_f} - \frac{m}{\rho_i} \right) = 4 \left[ \frac{1}{998} - \frac{1}{1000} \right]$$
Given P = 10⁵ Pa$P = 10^5 \text{ Pa}$:
W = 10⁵ × 4 × [ (1000 - 998)/(998 × 1000) ] = (8 × 10⁵)/(998 × 10³) ≈ 0.8 J$$W = 10^5 \times 4 \times \left[ \frac{1000 - 998}{998 \times 1000} \right] = \frac{8 \times 10^5}{998 \times 10^3} \approx 0.8 \text{ J}$$
Step 1: Internal Energy Calculation
Using the First Law of Thermodynamics:
Δ U = Q - W$$\Delta U = Q - W$$
Δ U = 268800 - 0.8 = 268799.2 J$$\Delta U = 268800 - 0.8 = 268799.2 \text{ J}$$
Pattern Recognition
When heating solids or liquids, the expansion is so small that Δ V$\Delta V$ is negligible, meaning W ≈ 0$W \approx 0$, making Δ U ≈ Δ Q$\Delta U \approx \Delta Q$. Subtracting the exact 0.8J$0.8\text{J}$ provides the highly precise answer required here.
Chapter Mix
Class 11 Physics: Thermodynamics
Class 11 Physics: Thermal Properties of Matter