What is the volume of a lead ball at – 12°C if volume at 160°C. is 530 cm³?

To determine the volume of a lead ball at –12°C, given its volume at 160°C is 530 cm³, we can use the concept of thermal expansion. Materials, including metals like lead, expand when heated and contract when cooled. The relationship between temperature change and volume change can be expressed using the formula for volumetric thermal expansion.

Understanding Thermal Expansion

Thermal expansion can be described by the formula:

ΔV = V₀ × β × ΔT

Where:

• ΔV = change in volume
• V₀ = original volume
• β = coefficient of volumetric expansion for the material
• ΔT = change in temperature

Coefficient of Volumetric Expansion for Lead

The coefficient of volumetric expansion (β) for lead is approximately 0.000018 per °C. This value indicates how much the volume of lead changes with temperature.

Calculating the Change in Volume

First, we need to calculate the temperature change (ΔT) from 160°C to –12°C:

ΔT = T_final - T_initial = -12°C - 160°C = -172°C

Now, we can substitute the values into the volumetric expansion formula:

ΔV = V₀ × β × ΔT

ΔV = 530 cm³ × 0.000018/°C × (-172°C)

Performing the Calculation

Calculating ΔV:

ΔV = 530 cm³ × 0.000018 × -172

ΔV = 530 × -0.003096 = -1.644 cm³

Finding the Volume at –12°C

Now, we can find the new volume (V_final) at –12°C:

V_final = V₀ + ΔV

V_final = 530 cm³ - 1.644 cm³ = 528.356 cm³

Thus, the volume of the lead ball at –12°C is approximately 528.36 cm³.