Question

The heat of fusion of water is 6.01 kJ/mol. The heat capacity of liquid water is 75.3 J/mol · K. The conversion of 50.0 g of ice at 0.00°C to liquid water at 22.0°C requires __________ kJ of heat.

Answer 1

Answer:

$21.5$ kJ

Explanation:

$M$ = mass of  1 mole of water = 18 g

$m$ = given mass of water = 50 g

Number of moles of water in the given amount is given as

$n=\frac{m}{M}$

$n=\frac{50}{18}$

$n=2.8$

$L$ = Heat of fusion of water = 6.01 kJ/mol = 6010 J/mol

$c$ = specific heat of water = 75.3 J/mol

$\Delta T$ = Change in temperature = 22.0 °C

Amount of heat required is given as

$Q = nL + n c \Delta T$

$Q = (2.8) (6010) + (2.8) (75.3) (22)$

$Q = 21466.5$ J

$Q = 21.5$ kJ