A drilling machine is used to drill a hole in a 10 kg metal block. Power of the drilling machine is 1 KW and it is used for 5 minutes. If 40% of the work done by the drilling machine is converted into heat in the metal block, the rise in its temperature is (specific heat - 0.12 cal $g{m}^{-1}$ ${}^{0}C^{-1}$ and J = 4.2 J/Cal)

Complete Solution:

Given Data:

• Mass of the metal block (m) = 10 kg = 10,000 g
• Power of the drilling machine (P) = 1 kW = 1000 J/s
• Time (t) = 5 minutes = 5 × 60 = 300 s
• Fraction of work converted into heat = 40% = 0.4
• Specific heat capacity (s) = 0.12 cal/g/°C
• Conversion factor (J) = 4.2 J/cal

The total work done by the drilling machine is:

W = P × t

Substituting values:

W = 1000 × 300 = 300,000 J

Since 40% of the work is converted into heat:

Q = 0.4 × W

Substituting values:

Q = 0.4 × 300,000 = 120,000 J

Convert Q to calories:

Q = 120,000 ÷ 4.2 = 28,571.43 cal

The heat absorbed by the metal block is:

Q = m × s × ΔT

Rearrange to solve for ΔT:

ΔT = Q / (m × s)

Substituting values:

ΔT = 28,571.43 / (10,000 × 0.12)

ΔT = 28,571.43 / 1,200

ΔT = 25°C

Final Answer:

The rise in temperature of the metal block is: 25°C