Calculate the quantity of heat needed to change the temperature of 60g of ice at 0 °C to 80 °C. [specific…

Analyze the question: The ice at 0 °C must first melt into water at 0 °C (latent heat), and then the water is heated from 0 °C to 80 °C (specific heat capacity). The total heat is the sum of both.

Concept:

• Heat to melt ice: Q₁ = m × L
• Heat to warm water: Q₂ = m × c × Δθ
• Total: Q = Q₁ + Q₂

Solution:

Mass: m = 60 g = 0.06 kg

1. Melting the ice: Q₁ = 0.06 × 3.36 × 10⁵ = 20,160 J = 20.16 kJ

2. Heating the water from 0 °C to 80 °C: Q₂ = 0.06 × 4.2 × 10³ × 80 = 20,160 J = 20.16 kJ

3. Total heat: Q = 20.16 + 20.16 = 40.32 kJ

Answer: The total is 40.32 kJ, which is option C.

Tip: Whenever a question starts with ice at 0 °C and ends with warm water, remember there are two stages: first melting (no temperature change, use mL), then heating (use mcΔθ). Students who forget the first stage end up with 20.16 kJ (option B), which is a common trap. Always check whether a phase change is involved.