Question

One end of a steel rod (K = 46 J s−1 m−1°C−1) of length 1.0 m is kept in ice at 0°C and the other end is kept in boiling water at 100°C. The area of cross section of the rod is 0.04 cm2. Assuming no heat loss to the atmosphere, find the mass of the ice melting per second. Latent heat of fusion of ice = 3.36 × 105 J kg−1.

Answer 1

$\huge\bold\star{\underline{\underline{\mathcal\purple{Nmste}}}}\star$

$<b>$

Solution :

K=46J/m−s−∘C,l=1m,

A=0.04cm2=4×10−6m2

QT=kA(T1−T2)l

=46×4×106×(100−0)1

=184×(10−4)

mL==184×(10−4)

m=(184×10−4)/Lm=(184xx10^(-4))/(3.36xx10^(5))=5.5xx10^(-5)g.`

▌│█║▌║▌║▌│█║▌║▌║♥️

Answer 2

$\huge{\underline{\underline{\mathfrak{Thank You}}}}$

K=46J/m−s−∘C,l=1m,

A=0.04cm2=4×10−6m2

QT=kA(T1−T2)l

=46×4×106×(100−0)1

=184×(10−4)

mL==184×(10−4)

m=(184×10−4)/Lm=(184xx10^(-4))/(3.36xx10^(5))=5.5xx10^(-5)g.`

$\huge{\underline{\underline{\mathfrak{Thank You}}}}$