The specific latent heat of vapourization of lead is 8.7 × $10^5 Jkg_1$. This means that 8.7 × $10^5$ J of thermal energy is required to change

Analyze the question: The question asks what it means physically for lead to have a specific latent heat of vaporization of 8.7 × 10⁵ J/kg.

Concept: Specific latent heat of vaporization is the quantity of heat energy required to change 1 kg of a substance from liquid to vapour (gas) at constant temperature (its boiling point), without a change in temperature.

Formula: Q = mL, so L = Q/m, with units J/kg.

Solution: Take each option in turn:

• A. "A given mass" is too vague. Specific latent heat refers to a unit mass (1 kg), so A is wrong.
• B. 1 kg of molten (liquid) lead changing to vapour at constant temperature and pressure. This matches the definition exactly, so B is correct.
• C. It starts from solid lead. Going from solid to vapour would involve melting first (and the latent heat of fusion), so C is wrong.
• D. "For every degree rise in temperature" describes specific heat capacity, not latent heat. During a change of state the temperature stays constant, so D is wrong.

Here 8.7 × 10⁵ J is needed to turn 1 kg of liquid lead into vapour at its boiling point, which is what B says.

Answer: B

Exam tip: Remember the difference this way:

• Specific heat capacity is energy per kg per degree (J/kg·K), and the temperature changes.
• Specific latent heat is energy per kg (J/kg), and the temperature stays constant during the change of state.

If you see "1 kg" and "constant temperature" together, it's latent heat. Watch the starting state too: vaporization starts from liquid, fusion starts from solid.