Question

ᴀ 10 ᴋᴡ ᴅʀɪʟʟɪɴɢ ᴍᴀᴄʜɪɴᴇ ɪs ᴜsᴇᴅ ғᴏʀ 5 ᴍɪɴᴜᴛᴇs ᴛᴏ ʙᴏʀᴇ ᴀ ʜᴏʟᴇ ɪɴ ᴀɴ ᴀʟʟᴜᴍɪɴɪᴜᴍ ʙʟᴏᴄᴋ ᴏғ ᴍᴀss 10 × 10^3 ᴋɢ.ɪғ 40% ᴏғ ᴛʜᴇ ᴡᴏʀᴋ ᴅᴏɴᴇ ɪs ᴜᴛɪʟɪsᴇᴅ ᴛᴏ ʀᴀɪsᴇ ᴛʜᴇ ᴛᴇᴍᴘᴇʀᴀᴛᴜʀᴇ ᴏғ ᴛʜᴇ ʙʟᴏᴄᴋ , ғɪɴᴅ ᴛʜᴇ ʀɪsᴇ ɪɴ ᴛᴇᴍᴘᴇʀᴀᴛᴜʀᴇ ᴏғ ᴛʜᴇ ᴀʟʟᴜᴍɪɴɪᴜᴍ ʙʟᴏᴄᴋ ?
( sᴘᴇᴄɪғɪᴄ ʜᴇᴀᴛ ᴏғ ᴀʟʟᴜᴍɪɴɪᴜᴍ = 0.9 ᴊ/ᴋɢ/ᴋ )

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Answer 1

Answer:

Heya sathvika ♥️

Explanation:

✏️Here, P=10kW=104W,m=8kg 

time, t=2.5minute=2.5×60=150s

Specific heat, s=0.91Jg−1C−1

Total energy =P×t=104×150=15×105J

As 50% of energy is lost,

∴ Energy available, ΔQ=21×15×105=7.5×105J

As ΔQ=msΔT 

ΔT=msΔQ=8×103×0.917.5×105≈103

✒️Hope it helps ♥️☺️✌️...

Answer 2

Work done by drilling machine in 5 min

We know that,

$\boxed {{ \rm{w = power \times time}}}$

$\rm = 10 \times 10 {}^{3} \times 5 \times 60$

$= 3 \times 10 {}^{6} j$

The energy utilized to rise the temperature of the block

$\boxed{\textsf{40 \% of w}}$

$\rm = 3 \times 10 {}^{6} \times \large \frac{40}{100}$

$\rm = 12 \times 10 {}^{5} j$

Now heat gained by aluminum

=mass ×specific heat×increase in temperature

$\rm12 \times 10 {}^{5} = (10 \times 10 {}^{3} ) \times 0.9 \times \Delta \: t$

$\therefore \Delta \: t = \large \frac{12 \times 10 {}^{5} }{0.9 \times 10 {}^{4} }$

$\boxed{ \rm = 133.3 {}^{0} c}$