A metal rod of length 10 cm and area of cross section `2.8 times 10^-4 m^2` is covered with a non-conducting substance. One end of it is maintained at `80^@C`, while the other end is put in ice at `0^@C`. It is found that 20 gm of ice melts of 5 min. The thermal conductivity of the metal in `Js^-1 m^-1 K^-1` is (Latent heat of ice is `80 cal g^-1`).

Options

A. 70
B. 80
C. 90
D. 100

Explanation

Given,
length of rod , l=10 cm =0.1 m
area of cross section of rod `A=2.8 times 10^-4 m^2`
temperature at one end, `T_1=80^@C`
temperature at other end , `T_2=0^@C`
quantity of melted ice ,m= 20 gm
time taken to melt ice, t=5 min =300sec
and latent heat of ice , `s=80 cal g^-1`
Now, rate of the heat flow `=(m times s times 4184)/t`
`=(20 times 80 times 4.184)/300=22.314 J//s`
Rate of heat flow in the rod, `DeltaQ=(kA.Delta T)/l`
`22.314=(k(2.8 times10^-4) times80)/0.1`
`therefore k=99.61 approx 100 J s^-1 m^-1 K^-1`