How much heat is required to convert 8.0 g of ice at `-15^@C` to steam at `100^@C`? (Given, `c_(ice) = 0.53 cal//g.^@C, L_f = 80 cal//g and L_v = 539 cal//g, `
and `c_(water) = 1 cal//g.^@C)` .

Explanation

`Q_(1) = mc_(ice)(T_(f)-T_(i))`
`(8.0)(0.53)[0-(-15)]`=63.6 cal
`Q_(2)` = mL_(f)=(8)(80)=640cal
`Q_(3)=mc_(water)(T_(f)-T_(i))`
(8.0)(1.0)[(100-0]=800 cal
`Q_(4)=mL_(V)`=(8.0)(539)=4312 cal
`therefore` Net heat required
`Q=Q_(1)+Q_(2)+Q_(3)+Q_(4)`=(5815.6)cal