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How much heat is required to convert 8.0 g of ice at `-15^@` to steam at `100^@`? (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)` .

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Correct Answer - A::C
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`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` .

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