Solveeit Logo
Login

Question

Question: The amount of heat required to reach the outer skin of aluminium to its boiling point is...

The amount of heat required to reach the outer skin of aluminium to its boiling point is

A

129.7 J

B

140.7 J

C

196.4 J

D

554.7 J

Answer

129.7 J

Explanation

Solution

The temperature change required is ΔT=2500∘C−(−50∘C)=2550∘C\Delta T = 2500^\circ C - (-50^\circ C) = 2550^\circ C. The volume of aluminium heated by the laser spot (1 mm diameter, so 0.5 mm radius or 0.05 cm radius) through the 3 cm thick skin is V=πr2d=π(0.05 cm)2(3 cm)=0.0075π cm3V = \pi r^2 d = \pi (0.05 \text{ cm})^2 (3 \text{ cm}) = 0.0075\pi \text{ cm}^3. The mass of this volume is m=ρV=(2.4 g/cm3)×(0.0075π cm3)=0.018π gm = \rho V = (2.4 \text{ g/cm}^3) \times (0.0075\pi \text{ cm}^3) = 0.018\pi \text{ g}. The heat required is Q=mcΔT=(0.018π g)×(0.9 J/g°C)×(2550∘C)=41.318π JQ = mc\Delta T = (0.018\pi \text{ g}) \times (0.9 \text{ J/g°C}) \times (2550^\circ C) = 41.318\pi \text{ J}. Using π≈3.14\pi \approx 3.14, Q≈129.7 JQ \approx 129.7 \text{ J}.