Correct Answer - A::C
`Q_1 = mc_(ice) (T_f - T_i)`
`(8.0)(0.53)[0-(-15)] = 63.6 cal`
`Q_2 = mL_f = (8)(80) = 640 cal`
`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`
`:.` Net heat required, `Q = Q_1 + Q_2+Q_3+Q_4`
= `5815.6 cal` .