Question

The heat capacity of liquid water is 75.6j/molk, while the enthalpy of fusion of ice is 6.0kj/mol. What is the smallest number of ice cubes at 0oc, each containing 9.0 g of water, needed to cool 500 g of liquid water from 20oc to 0oc?

Answer 1

The smallest number of cubes is 14.

Given:

Heat capacity of liquid water = 75.6 J/mol.K

Enthalpy of fusion of ice = 6.0 kJ/mol = 6000 J/mol

Mass of ice cubes = 9 g

Mass of liquid water = 500 g

To find:

Number of cubes = ?

Formula used:

Heat Transfer:

$\bold{Q=mcT}$

Where,

m = Mass

c = Heat capacity

T = Temperature change

Solution:

Water (Heat lost):

$\bold{Q=mcT}$

$\bold{Q=\frac{500}{18}\times75.6\times20}$

Ice (Heat gain):

$\bold{Q=mcT}$

$\bold{Q= \frac{9}{18}\times n\times6000}$

Heat lost = Heat gain

$\bold{\frac{500}{18}\times75.6\times20= \frac{9}{18}\times n\times6000}$

$\bold{\therefore n = 14}$