An iron meteorite melts when it enters the Earth\'s atmosphere. If its Initial t

Question

An iron meteorite melts when it enters the Earth's atmosphere. If its Initial temperature was -105 degree C outs.de of Earth's atmosphere, calculate the minimum velocity the meteorite must have had before it enters Earth's atmosphere. 1S20 m/s 760 m/s 1250 m/s 3040 m/s A 126.5-g insulated aluminum cup at 18.00 degree C is filled with 132.5 g of water at 46.25 degree C. After a few minutes, equilibrium Is reached. Determine the final temperature. 32.13 degree C 34.91 degree C 16.2 5 degree C 41.44 degree c

Explanation / Answer

Question-19

Q= mciron* delta T +m Lvaporisation= m* 449* [Tf- Ti] +m* Lfusion

We have the common data :-

Tf is the melting point of ice= 15380C

ciron= 449 J/kg

Lfusion of iron=272 kJ/kg/K

Thus we get change in temperature as ,Tf-Ti= 15380C--1050C=16430C=1916.15 K

Q=m* 449*1916.15 K +m* 272 kJ/kg= m*1.13*10 6 J

Here energy is equal to Ke of the meteorite

Q= KE

Q= 1/2mv2

m*1.13*10 6 J= 1/2mv2

v2= 2*1.13*10 6

v= 1505 m/s------------------------Option a

Question-20

Let T be the final temperature

Q Aluminium=Q water

mAl * cAl* [Ti-T]= mwater * c water* [T-Tf]

126.5g*0.904J/g/K*[ (18+273)K-T]=132.5*g*4.187 *10-3J/g/K * [T- (46.25+273)]

126.5*0.904*[291K-T]= 132.5*4.187 *10-3* [T-319.25]

114.36*[291K-T]= 0.555*[T-319.25K]

0.206*[291K-T]=[T-319.25K]

59.98K-0.206T=T-319.25K

379.24K= 1.206T

T=314.46K= 41.450C

T=41.450C--------------------Option D

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