Question

What amount of ice must be added to 250 g of water at 15°c to cool the water to 0°c and have no ice left

Answer 1

Answer:

$\mathbf{m= 47 \ g}$

Explanation:

The heat needed to melt ice $(q_1)$ + heat needed to cool water $(q_2)$ = 0

$m\Delta H_{fusion} + mc \Delta T = 0$

$m \times 333.55 J g^{-1} + 250 \ g \times 4.184 \ Jg^{-1 \ 0 } \times (-15.0 \ ^0 C) = 0$

$m \times 333.55 J g^{-1} =15690$

$m=\dfrac{15690 \ J }{333.55 J g^{-1} }$

$\mathbf{m= 47 \ g}$