Question

Beryllium is a rare metal that is gray in color, strong, and lightweight. The following data was obtained from a calorimeter experiment:

5 sets of equations: m beryllium equals 10.1 grams. M water equals 38.8 grams. Delta T beryllium equals final T minus initial T equals 27.7 degrees Celsius minus 100 degrees Celsius. Delta T water equals Final T minus initial T equals 28.4 degrees Celsius minus 20.2 degrees Celsius. C water equals 4.18 joules per gram per degree Celsius.

The specific heat of beryllium is
J/g°C
C Subscript metal Baseline = StartFraction negative C Subscript water Baseline m Subscript water Baseline delta T Subscript water Over m Subscript metal Baseline delta T Subscript metal EndFraction. An irregular chunk of gray metal is shown.

Photo by Heinrich Pniok

Answer 1

Use the heat-balance formula you wrote:

C_metal = - (C_water m_water ΔT_water) / (m_metal ΔT_metal)

Given:
- m_metal = 10.1 g
- m_water = 38.8 g
- ΔT_metal = 27.7°C − 100.0°C = −72.3°C
- ΔT_water = 28.4°C − 20.2°C = 8.2°C
- C_water = 4.18 J/g·°C

Compute numerator: 4.18 × 38.8 × 8.2 = 1329.9088 J
Denominator: 10.1 × (−72.3) = −730.23 g·°C

C_metal = −(1329.9088) / (−730.23) = 1329.9088 / 730.23 ≈ 1.82 J/g·°C

So the specific heat of the beryllium sample is about 1.82 J/g·°C (≈1.82 J·g⁻¹·K⁻¹), which agrees well with the tabulated value (~1.825 J/g·K).