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.