`1 gm` of ice at `0^@C` is converted to steam at `100^@C` the amount of heat required will be `(L_("steam") = 536 cal//g)`.

Options

A. 100 cal
B. `0.01 cal^@C^(-1)`
C. 720 cal
D. 1 kilocal

Answer

C

Explanation

Using the formula,
1) `Q = m xx s xx DeltaT`,
where Q, s and `DeltaT` are heat required, specific heat capacity and change in temperature respectively.
2) `Q = mL `
where m and Lare mass and latent heat respectively.
Values of latent heat of fuşion and vapourization are 80 cal/g and 540 cal/g, while specific heat of water is `1 "cal / kg /" ""^@C`.
Total heat required must be
`= m xx 80 + m xx 1 xx 100 + m xx 540 `
`= m(720) = 1 720 = 720 cal `