A drilling machine of 5 kW is used to drill a hole in the block of copper of mass 4 kg . Calculate the riลe 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 โ 1 โ C โ I = 0.385 ๐ฝ ๐ โ 1 โ ๐ถ โ ๐ผ.
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
5000โJ/s5000 , text{J/s}) and it runs for 5 minutes (or
300โs300 , text{s}). The total energy supplied is:
Energyย supplied=PowerรTime
text{Energy supplied} = text{Power} times text{Time}
Energyย supplied=5000โJ/sร300โs=1,500,000โJ
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:
Energyย forย heating=0.75ร1,500,000=1,125,000โJtext{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=mโ cโ ฮTQ = m cdot c cdot Delta T
Where:
-
QQ
= Heat energy ( 1,125,000โJ1,125,000 , text{J}
)
-
mm
= Mass of the block ( 4โkg=4000โg4 , text{kg} = 4000 , text{g}
)
-
cc
= Specific heat capacity of copper ( 0.385โJ/g/ยฐC0.385 , text{J/g/ยฐC}
)
-
ฮTDelta T
= Temperature rise ( ??
)
Rearranging the formula to solve for
ฮTDelta T:
ฮT=Qmโ cDelta T = frac{Q}{m cdot c}
Substitute the values:
ฮT=1,125,0004000โ 0.385Delta T = frac{1,125,000}{4000 cdot 0.385}
ฮT=1,125,0001540โ730.52โยฐCDelta T = frac{1,125,000}{1540} approx 730.52 , text{ยฐC}
The rise in temperature of the copper block is approximately 730.52ยฐC.