1. **Problem statement:** (a) Define specific heat capacity, specific latent heat, and thermal equilibrium. (b) Calculate the final temperature when a 2.0 kg copper block at 100 °C is placed in 1.5 kg of water at 20 °C. (c) Given 0.2 kg ice at 0 °C added to 0.5 kg water at 30 °C, calculate heat to melt ice, heat lost by water cooling to 0 °C, and determine if all ice melts plus final temperature. (d) Explain kinetic theory differences between temperature and heat. 2. **(a) Definitions:** (i) Specific heat capacity is the amount of heat required to raise the temperature of 1 kg of a substance by 1 °C. (ii) Specific latent heat is the amount of heat energy required to change the state of 1 kg of a substance without changing its temperature. (iii) Thermal equilibrium occurs when two objects in contact reach the same temperature and no heat flows between them. 3. **(b) Final temperature calculation:** Let the final temperature be $T_f$ in °C. Heat lost by copper = Heat gained by water (no heat loss outside system) Heat lost by copper: $Q_c = m_c c_c (T_i - T_f) = 2.0 \times 390 \times (100 - T_f)$ Heat gained by water: $Q_w = m_w c_w (T_f - T_i) = 1.5 \times 4200 \times (T_f - 20)$ Set equal: $$2.0 \times 390 \times (100 - T_f) = 1.5 \times 4200 \times (T_f - 20)$$ Simplify: $$780 (100 - T_f) = 6300 (T_f - 20)$$ $$78,000 - 780 T_f = 6300 T_f - 126,000$$ Bring terms: $$78,000 + 126,000 = 6300 T_f + 780 T_f$$ $$204,000 = 7080 T_f$$ Solve for $T_f$: $$T_f = \frac{204,000}{7080} \approx 28.81\ ^\circ C$$ 4. **(c) Ice and water mixture:** (i) Heat to melt ice: $$Q_{melt} = m \times L_f = 0.2 \times 334,000 = 66,800\ \text{J}$$ (ii) Heat lost by warm water cooling to 0 °C: $$Q_{water} = m c \Delta T = 0.5 \times 4200 \times (30 - 0) = 63,000\ \text{J}$$ (iii) Compare heat required to melt ice with heat lost by water: Heat lost by water (63,000 J) < Heat needed to fully melt ice (66,800 J) Since water loses less heat than needed to melt all ice, not all ice melts. Calculate melted ice mass: $$melted\ ice = \frac{Q_{water}}{L_f} = \frac{63,000}{334,000} \approx 0.188 \text{ kg}$$ Remaining ice mass = 0.2 - 0.188 = 0.012 kg unfrozen. Final temperature = 0 °C (since system is partially melted ice and water in equilibrium). 5. **(d) Kinetic theory explanation:** Temperature measures the average kinetic energy of particles, indicating how fast they move. Heat is the total energy transferred due to temperature difference, depending on both particle energy and quantity of matter. Hence, temperature is an intensive property, heat is extensive. Final answers: (b) $\boxed{28.81\ ^\circ C}$ (c)(i) $66,800$ J (c)(ii) $63,000$ J (c)(iii) Not all ice melts; final temp is $0\ ^\circ C$.