Question

1 mol of water at 10°C is converted into ice at –10°C. The change in enthalpy for complete conversion is

[Given : C_p of water = x JK^–1 mol^–1 ]

C_p of ice = y JK^–1 mol^–1

^{\(\Delta\text{h}_{\text{fusion}} = Z J \)}

(1) (–10x – 10y – z) J

(2) (10x + 10y + z) J

(3) (x + y – z) J

(4) 10(x + y – z) J

Answer 1

Correct option is : (1) (–10x – 10y – z) J

\(\underset{283 \ k}{H_2O(l)} \rightarrow \underset{273 \ k}{ H_2O(l)}\)

\(\Delta H_1 = nC_p\Delta T\)

\(= 1 × (10) = –10x J\)

\(\underset {273 \ k}{H_2O(l)} \rightarrow \underset{273 \ k}{H_2O(s)}\)

\(\Delta H_ {fusion} = ZJ\)

\(\underset {273 \ k}{H_2O(s)} \rightarrow \underset{263 \ k}{H_2O(l)}\)

\(\Delta H_2 = y (-10) J\)

\(= –10y J\)

\(\Delta H_{net} = \Delta H_1 + \Delta H_{fusion} + \Delta H_2\)

\(= –10x – z – 10y\)