Latent heat and calorimetry

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The added energy goes into **breaking the bonds** between particles (latent heat), not into their kinetic energy — so the temperature does not change.

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The **energy needed to change the state of 1 kg** of a substance with **no temperature change**. Unit: J kg⁻¹.

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$Q = mL$ — energy = mass × specific latent heat. **Given** in the data booklet. Used for the flat parts (state change).

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$Q = mc\Delta T$ — mass × specific heat capacity × temperature change. **Given** in the data booklet. Used for the sloping parts.

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**Fusion (Lf)** = melting/freezing. **Vaporisation (Lv)** = boiling/condensing. For one substance, **Lv ≫ Lf**.

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A **state change** (melting or boiling) at **constant temperature** — use $Q = mL$.

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The **temperature is changing** (warming or cooling) — use $Q = mc\Delta T$.

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Vaporising fully separates the particles, needing far more energy than melting (Lv ≫ Lf), so it takes longer at a steady heating rate.

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**Energy lost by the hot object = energy gained by the cold object.** Add one Q-term per step (warm, melt, warm…).

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Use a **separate Q-term for each step**: $Q = mc\Delta T$ for each temperature change and $Q = mL$ for each state change, then add them.

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Some thermal energy is **lost to the surroundings** or absorbed by the **container**, which the ideal 'no losses' calculation ignores.

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$Q = mL = 0.50 \times 3.3\times10^{5} = 1.65\times10^{5}$ J (≈ 1.7 × 10⁵ J).