The opposite faces of a cubical block of iron of cross-section `4` square cm are kept in contact with steam and melting ice. Calculate the quantity of ice melted at the end of `10` minutes. `k` for iron `= 0.2 CGS` units.

Options

A. `300 g`
B. `150 g`
C. `75 g`
D. `450 g`

Answer

A

Explanation

Let `x` be each side of the cube.
As area `(A)` of each face is `4cm^(2)`
`A = x^(2) = 4 cm^(2) or x=2 cm`
Further, with usual notation,
`(T_(1)-T_(2)) = (100-0) = 100^(@)C`
`(T_(1)` = temperature of steam, `T_(2)` = temperature of melting ice)
`k=0.2 CGS` unit , `t=10 min = 10xx60s = 600s`
As, `Q = (KA(T_(1)-T_(2))t)/(x)`
`Q = (0.2 xx 4 xx 100 xx 600)/(2) = 24,000 cal`
if `m` gram of ice be melting with this heat and `L` be the latent heat of fusion of ice, `Q =mL`
Or, `m=(Q)/(L) = (24000)/(80) (as L = 80cal//g)`
or, `m=300g`.