0.50 kg of water at 75 °C is poured into 1.5 kg of water at 15 °C in an insulated jug…

Step 1. Final temperature. $\dfrac{(0.50 \times 75 + 1.5 \times 15)\ \text{kg}\,^\circ\text{C}}{2.0\ \text{kg}} = 30\ ^\circ\text{C}$, or $303.15\ \text{K}$.

Step 2. Hot water. $0.50\ \text{kg} \times 4186\ \text{J kg}^{-1}\text{K}^{-1} \times \ln\dfrac{303.15\ \text{K}}{348.15\ \text{K}} = -289.7\ \text{J/K}$.

Step 3. Cold water. $1.5\ \text{kg} \times 4186\ \text{J kg}^{-1}\text{K}^{-1} \times \ln\dfrac{303.15\ \text{K}}{288.15\ \text{K}} = +318.6\ \text{J/K}$.

Step 4. Total. $-289.7\ \text{J/K} + 318.6\ \text{J/K} = 29.0\ \text{J/K}$.