Work done in converting one gram of ice at `-10^(@)C` into steam at `100^(@)C` is

Options

A. 3045 J
B. 6056 J
C. 721 J
D. 6 J

Answer

A

Explanation

Work done in converting 1 g of ice at `-10^@C` to steam at `100^@C`
= Heat supplied to raise temperature of 1 g of ice from `10^@C` to `0^@C(mxxc_("ice")xxDeltaT)` + Heat supplied to convert 1 g ice into water at `0^@C (mxxL_("ice"))`
+ Heat supplied to raise temperature of 1 g of water from
`0^@C` to `100^@C(mxxc_("water")xxDeltaT)`
+ heat supplied to convert 1 g water into steam at `100^@C [mxxL_("vapour")]`
`=[mxxc_("ice")xxDeltaT]+{mxxL_("ice")]+[mxxc_("water")xxDeltaT]+[mxxL_("vapour")]`
`=[1xx0.5xx10]+[1xx80]+[1xx1xx100]+[1xx540]=725cal=725xx4.2=3045 J`