A drilling machine of 5 kW is used to drill a hole in the block of copper of mass 4 kg . Calculate the rişe in temperature of the block in 5 minutes.if 75 % 75 % of the energy is used in heating the block. Given specific heat of copper = 0.385 J g − 1 ∘ C − I = 0.385 𝐽 𝑔 − 1 ∘ 𝐶 − 𝐼

To calculate the rise in temperature of the block, we can follow these steps:

Step 1: Calculate the total energy supplied

The power of the drilling machine is 5 kW (or

5000 J/s5000 , text{J/s}) and it runs for 5 minutes (or

300 s300 , text{s}). The total energy supplied is:

Energy supplied=Power×Time

text{Energy supplied} = text{Power} times text{Time}

Energy supplied=5000 J/s×300 s=1,500,000 J

text{Energy supplied} = 5000 , text{J/s} times 300 , text{s} = 1,500,000 , text{J}

Step 2: Calculate the energy used to heat the block

Only 75% of the energy is used to heat the block. So, the energy used for heating is:

Energy for heating=0.75×1,500,000=1,125,000 Jtext{Energy for heating} = 0.75 times 1,500,000 = 1,125,000 , text{J}

Step 3: Use the formula for heat transfer to calculate the temperature rise

The formula for heat transfer is:

Q=m⋅c⋅ΔTQ = m cdot c cdot Delta T

Where:

• QQ

  = Heat energy ( 1,125,000 J1,125,000 , text{J}

  )

• mm

  = Mass of the block ( 4 kg=4000 g4 , text{kg} = 4000 , text{g}

  )

• cc

  = Specific heat capacity of copper ( 0.385 J/g/°C0.385 , text{J/g/°C}

  )

• ΔTDelta T

  = Temperature rise ( ??

  )

Rearranging the formula to solve for

ΔTDelta T:

ΔT=Qm⋅cDelta T = frac{Q}{m cdot c}

Substitute the values:

ΔT=1,125,0004000⋅0.385Delta T = frac{1,125,000}{4000 cdot 0.385}

ΔT=1,125,0001540≈730.52 °CDelta T = frac{1,125,000}{1540} approx 730.52 , text{°C}

The rise in temperature of the copper block is approximately 730.52°C.