A solid sample of lead with a mass of 0.80\text{ kg} is kept at its melting point of 327^\… — JAMB UTME

A solid sample of lead with a mass of $0.80\text{ kg}$ is kept at its melting point of $327^\circ\text{C}$ . If $15,000\text{ J}$ of thermal energy is supplied to the sample, calculate the mass of lead, in kilograms, that remains in the solid state. (Take the specific latent heat of fusion of lead as $2.5 \times 10^4\text{ J/kg}$ ).

Answer: 0.2 kg

Answer

The mass of lead remaining in the solid state is

0.20\text{ kg}

Thermal energy

Q = 15,000\text{ J}

supplied to lead at its melting point melts a portion calculated by

m_{\text{melted}} = \frac{Q}{L_f} = \frac{15,000}{25,000} = 0.60\text{ kg}

. Subtracting this melted mass from the original

0.80\text{ kg}

yields

0.20\text{ kg}

of remaining solid lead.

Step-by-Step Solution

Calculate the mass of lead that melts.

Melted mass

m_{\text{melted}} = 0.60\text{ kg}

At the melting point, thermal energy supplied goes entirely into phase change without changing temperature:

Q = m_{\text{melted}} L_f

Determine the remaining mass of solid lead.

Remaining solid mass

m_{\text{solid}} = 0.20\text{ kg}

The un-melted portion equals the initial total mass minus the mass that has melted (

m_{\text{solid}} = m_{\text{total}} - m_{\text{melted}}

Key Concept

Latent Heat of Fusion and Phase Change

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