a metal rod of length 10 cm and area of cross section 2.8 ×10 ^-4 m^2 is covered with a non conducting substance. one end of it us maintained at 80,while the other end is put in ice at 0.it is found, that 20 g of ive melts in 5 min. the thermal conductivity of the metal in Js^-1 m^-1 k^-1 is
Answers
A metal rod of length 20 cm and diamter 2 cm is convered with a non conducting substance. One of its ends is maintained at, while the other end is put at. It is found that 25 g ice melts in 5 min. calculate the coefficient of thermal conductivity of the metal. Latent that of ice =.
Answer: 100 J/S m K
Explanation:Given,20g of ice melting in 5 min i.e 300sec.....(1)
Latent heat of ice =80 cal/g.
Now,Heat transfered(Q)=m(ice) *Latent heat of ice.
Q=(20*10^-2Kg)(80*4.18J). {1cal=4.18J}....
Q=6.698*10^3........(2)
Therefore,Q/t=KA(Ti-Tf)/x
t=5min=300sec, A=2.8*10^-4 m^2. Ti=80*C,Tf=0*C,x=10cm=10*10^-2m.
6.698*10^3/300sec =K(2.8*10^-4)(80-0)/10*10^-2.
=)K =6.698*10^3/67.2
K=~~100 J/s m K.