Question

Determine the heat needed to raise \(60 \mathrm{~g}\) of \(\mathrm{Pb}\) from \(20^{\circ} \mathrm{C}\) to \(360^{\circ} \mathrm{C}\), given its specific heats are \(0.0306\) (solid) and \(0.0375\) (liquid). Its heat of fusion is \(5.86 \mathrm{cal} / \mathrm{g}\); its melting point is \(327^{\circ} \mathrm{C}\).

Short answer

The heat needed to raise 60g of Pb from 20°C to 360°C is calculated in 3 parts: heating the solid Pb, melting it, and then heating the liquid Pb. The total heat required is the sum of these three heats: - \(Q_s = (60 \times 0.0306) \times 307\) - \(Q_m = (60 \times 5.86)\) - \(Q_l = (60 \times 0.0375) \times 33\) To find the total heat needed, sum the calculated heats from the three stages: \[Q_{total} = Q_s + Q_m + Q_l\]

Step-by-step solution

Calculate heat needed to heat the solid Pb from 20°C to 327°C

(temperature change in solid phase) First, we need to find the heat required to heat up the solid Pb from 20°C to its melting point of 327°C. We will use the formula: \[Q_s = mc_sΔT_s\] Where: - \(Q_s\) = heat needed to heat the solid Pb to its melting point - m = mass (60g) - \(c_s\) = specific heat of solid Pb (0.0306 cal/g°C) - \(ΔT_s\) = change in temperature of solid Pb (327°C - 20°C) The temperature change is: \(ΔT_s\) = 327°C - 20°C = 307°C Now we can calculate the heat needed (in calories) to heat up the solid Pb: \(Q_s = (60 \times 0.0306) \times 307\)

Calculate the heat needed to melt the Pb

(heat of fusion) Next, we need to find the heat required to melt the solid Pb at 327°C. We will use the formula: \[Q_m = m \times H\] Where: - \(Q_m\) = heat needed to melt the Pb - m = mass (60g) - H = heat of fusion (5.86 cal/g) Now we can calculate the heat needed (in calories) to melt the Pb: \(Q_m = (60 \times 5.86)\)

Calculate heat needed to heat the liquid Pb from 327°C to 360°C

(temperature change in liquid phase) Finally, we need to find the heat required to heat up the liquid Pb from the melting point of 327°C to 360°C. We will use the formula: \[Q_l = mc_lΔT_l\] Where: - \(Q_l\) = heat needed to heat the liquid Pb to 360°C - m = mass (60g) - \(c_l\) = specific heat of liquid Pb (0.0375 cal/g°C) - \(ΔT_l\) = change in temperature of liquid Pb (360°C - 327°C) The temperature change is: \(ΔT_l\) = 360°C - 327°C = 33°C Now we can calculate the heat needed (in calories) to heat up the liquid Pb: \(Q_l = (60 \times 0.0375) \times 33\)

Combine the heats for all three stages

(total heat needed) Now that we have calculated the heat needed for each stage of the process, we can find the total heat needed by adding them all together. \[Q_{total} = Q_s + Q_m + Q_l\] After completing Steps 1, 2, and 3, we found: - \(Q_s = (60 \times 0.0306) \times 307\) - \(Q_m = (60 \times 5.86)\) - \(Q_l = (60 \times 0.0375) \times 33\) Plug these values into the formula for total heat (Q_total), and calculate the result.

Key concepts

Specific Heat

Specific heat is an important concept in chemistry used to describe how much heat energy is required to raise the temperature of a given mass of a substance by one degree Celsius. It is represented by the term 'c' in the heat formula. Specific heat varies between different substances. This means some materials will heat up faster than others, given the same amount of energy.

For lead (Pb), specific heat values are different when it is in solid or liquid form. In this exercise, the specific heat of solid lead is given as \(0.0306 \mathrm{cal} / \mathrm{g}^{\circ} \mathrm{C}\) and the specific heat of liquid lead is \(0.0375 \mathrm{cal} / \mathrm{g}^{\circ} \mathrm{C}\). This demonstrates that liquid Pb requires slightly more energy per gram to change its temperature, compared to its solid state.

Heat of Fusion

The heat of fusion is the energy required to change a substance from solid to liquid at its melting point. For lead, this value is significant because it determines how much energy is needed to convert lead from solid to liquid without changing its temperature.

In our exercise, the heat of fusion for lead is given as \(5.86 \mathrm{cal} / \mathrm{g}\). Understanding this helps in calculating how much heat energy is needed for the phase change when lead reaches its melting point. This value is crucial in processes where melting and solidification need to be controlled, like in metallurgical applications.

Temperature Change

Temperature change is a key factor in calculating heat energy. It involves determining the difference between the initial and final temperatures of a substance.

In the exercise, there are two temperature changes to consider: the heating of solid lead from \(20^{\circ}\mathrm{C}\) to its melting point at \(327^{\circ}\mathrm{C}\), and the heating of liquid lead from \(327^{\circ}\mathrm{C}\) to \(360^{\circ}\mathrm{C}\). Each temperature change requires careful calculation to find out the total heat required.

• For the solid phase: \(ΔT_s = 327^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C} = 307^{\circ} \mathrm{C}\)
• For the liquid phase: \(ΔT_l = 360^{\circ} \mathrm{C} - 327^{\circ} \mathrm{C} = 33^{\circ} \mathrm{C}\)

These calculations are crucial in understanding how substances respond to heat.

Melting Point

The melting point of a substance is the temperature at which it changes from solid to liquid. This is a critical point in calculations involving phase changes.

Lead has a melting point of \(327^{\circ} \mathrm{C}\). At this temperature, lead will transition from solid to liquid, absorbing energy without having its temperature raised further. Knowing the melting point helps define the phases in which heat energy needs to be calculated separately. This ensures accurate comprehension of the lead's behavior during heat application, particularly in heating and melting applications.

Lead (Pb)

Lead (Pb) is a dense metallic element with a high atomic number. It is known for its soft, malleable nature, and low melting point among metals. These properties make it highly versatile in industrial and chemical processes.

Lead is significant in exercises involving heat because its distinct physical properties directly affect how it responds to heat energy. Its lower specific heat values imply it requires less energy for temperature changes compared to other metals. Understanding lead's characteristics is vital when performing thermal calculations, as it informs us of energy distribution during heating, melting, and boiling processes. Additionally, the knowledge of lead's specific heat and heat of fusion aids in efficient thermal management in metallurgical and manufacturing processes.