**xviii.** Calculate the total heat required (a) to melt 180 g of ice at 0 $^0$C, (b) heat it to 100 $^0$C and then (c) vapourise it at that temperature. Given $\Delta_{fus} H^0$(ice) = 6.01 kJ mol$^{-1}$ at 0 $^0$C, $\Delta_{vap} H^0$(H$_2$O) = 40.7 kJ mol$^{-1}$ at 100 $^0$C specific heat of water is 4.18 J g$^{-1}$ K$^{-1}$ Ans. : (542.3 kJ)

**xviii.** Calculate the total heat required (a) to melt 180 g of ice at 0 ⁰C, (b) heat it to 100 ⁰C and then

Answer the following questions · Q18

Q. xviii. Calculate the total heat required

(a) to melt 180 g of ice at 0

^0

(b) heat it to 100

^0

C and then

\Delta_{fus} H^0

(ice) = 6.01 kJ mol

^{-1}

at 0

^0

\Delta_{vap} H^0

_2

O) = 40.7 kJ mol

^{-1}

at 100

^0

C specific heat of water is 4.18 J g

^{-1}

^{-1}

Ans. : (542.3 kJ)

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Step 1. Moles of water = 180 g / 18 g mol-1 = 10 mol.

Step 2 (melt). Heat to melt: 10 mol x 6.01 kJ/mol = 60.1 kJ.

Step 3 (heat 0 to 100 degC). Q = m c ΔT = 180 g x 4.18 J g-1 K-1 x 100 K = 75,240 J = 75.24 kJ. …