Correct Answer - A
Ice `(-10^(@)`C converts into steam as follows
`c_(i)` = Specific heat of ice, `c_(w)`= Specific heat of water)
Total heat required Q= `Q_(1) + Q_(2) + Q_(3) + Q_(4)`
`rArr Q= 1 xx 0.5(10) + 1 xx 80 + 1 xx 1xx(100-0) + 1 xx 540`
=725 cal
Hence, work done W = JQ = `4.2 xx 725` = 3045` J
