Question

$\huge{\underline{\underline{\mathfrak\green{hola}}}}$
Give expression for time of the cooling of a body through a particular range of temprature.​

Answer 1

$\huge{\underline{\mathfrak\red{hola}}}$

Question:-

ɢɪᴠᴇ ᴇxᴘʀᴇssɪᴏɴ ғᴏʀ ᴛɪᴍᴇ ᴏғ ᴛʜᴇ ᴄᴏᴏʟɪɴɢ ᴏғ ᴀ ʙᴏᴅʏ ᴛʜʀᴏᴜɢʜ ᴀ ᴘᴀʀᴛɪᴄᴜʟᴀʀ ʀᴀɴɢᴇ ᴏғ ᴛᴇᴍᴘʀᴀᴛᴜʀᴇ.

Answer:-

•)ɴᴇᴡᴛᴏɴ's ʟᴀᴡ ᴏғ ᴄᴏᴏʟɪɴɢ :-

• ɴᴇᴡᴛᴏɴ's ʟᴀᴡ ᴏғ ᴄᴏᴏʟɪɴɢ sᴛᴀᴛᴇs ᴛʜᴀᴛ ᴛʜᴇ ʀᴀᴛᴇ ᴏғ ʜᴇᴀᴛ ʟᴏss ᴏғ ᴀ ʙᴏᴅʏ ɪs ᴅɪʀᴇᴄᴛʟʏ ᴘʀᴏᴘᴏʀᴛɪᴏɴᴀʟ ᴛᴏ ᴛʜᴇ ᴅɪғғᴇʀᴇɴᴄᴇ ɪɴ ᴛʜᴇ ᴛᴇᴍᴘᴇʀᴀᴛᴜʀᴇs ʙᴇᴛᴡᴇᴇɴ ᴛʜᴇ ʙᴏᴅʏ ᴀɴᴅ ɪᴛs sᴜʀʀᴏᴜɴᴅɪɴɢs.

•) ғᴏʀᴍᴜʟᴀ :-

$\huge{\underline{\underline{\mathfrak\red{ϙ=ʜ.ᴀ.(ᴛ(t)-ᴛᵉⁿᵘ)}}}}$

• ϙ = ʀᴀᴛᴇ ᴏғ ʜᴇᴀᴛ ᴛʀᴀɴsғᴇʀ ᴏᴜᴛ ᴏғ ᴛʜᴇ ʙᴏᴅʏ
• ʜ = ʜᴇᴀᴛ ᴛʀᴀɴsғᴇʀ ᴄᴏᴇғғɪᴄɪᴇɴᴛ
• ᴀ = ʜᴇᴀᴛ ᴛʀᴀɴsғᴇʀ sᴜʀғᴀᴄᴇ ᴀʀᴇᴀ
• ᴛ = ᴛᴇᴍᴘᴇʀᴀᴛᴜʀᴇ ᴏғ ᴛʜᴇ ᴏʙᴊᴇᴄᴛ sᴜʀғᴀᴄᴇ
• ᴛ(ᴛ) = ᴛɪᴍᴇ ᴅᴇᴘᴇɴᴅᴇɴᴛ ᴛᴇᴍᴘᴇʀᴀᴛᴜʀᴇ
• ᴛᵉⁿᵘ = ᴛᴇᴍᴘᴇʀᴀᴛᴜʀᴇ ᴏғ ᴇɴᴠɪʀᴏɴᴍᴇɴᴛ

•) ɴᴇᴡᴛᴏɴ's ʟᴀᴡ ᴏғ ᴄᴏᴏʟɪɴɢ ᴄᴀʟᴄᴜʟᴀᴛᴏʀ

• ɪᴛ ɪs ᴇᴀsʏ ᴛᴏ ᴀᴘᴘʟʏ ɴᴇᴡᴛᴏɴ's ʟᴀᴡ ᴏғ ᴄᴏᴏʟɪɴɢ ᴡɪᴛʜ ᴏᴜʀ ᴄᴀʟᴄᴜʟᴀᴛᴏʀ. ᴊᴜsᴛ sᴘᴇᴄɪғʏ ᴛʜᴇ ɪɴɪᴛɪᴀʟ ᴛᴇᴍᴘᴇʀᴀᴛᴜʀᴇ (ʟᴇᴛ's sᴀʏ 100 °ᴄ), ᴛʜᴇ ᴀᴍʙɪᴇɴᴛ ᴛᴇᴍᴘᴇʀᴀᴛᴜʀᴇ (ʟᴇᴛ's sᴀʏ 22 °ᴄ) ᴀɴᴅ ᴛʜᴇ ᴄᴏᴏʟɪɴɢ ᴄᴏᴇғғɪᴄɪᴇɴᴛ (ғᴏʀ ᴇxᴀᴍᴘʟᴇ 0.015 1/s) ᴛᴏ ғɪɴᴅ ᴏᴜᴛ ᴛʜᴀᴛ ᴛʜᴇ ᴛᴇᴍᴘᴇʀᴀᴛᴜʀᴇ ᴅʀᴏᴘs ᴛᴏ 35 °ᴄ ᴀғᴛᴇʀ 2 ᴍɪɴᴜᴛᴇs.

Answer 2

Answer:

To find the time, set f (t) = 155 and solve for t. Now, let's take a look at the "official formula" for Newton's Law of Cooling. According to Newton's Law of Cooling, an object's temperature change rate is proportional to its own temperature and the temperature of the surrounding environment.

Explanation:

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