Heat energy of 184kJ is given to ice of mass 600g at - 12 ° C ,...

Given:

$∆ Q = 184 \times {10}^{3}J$,

$m = 0 . 600kg\text{at} - 12 ° C$,

$S = 2222 . 3J{kg}^{– 1} ° {C}^{– 1}$ and

$L = 336 \times {10}^{3}J{kg}^{- 1}$

Heat required to convert ice from $– 12 ° C$ to ice at $0 ° C$ will be, 

${Q}_{1} = 0 . 600 \times 2222 . 3 \times 12 = 16000 . 56J$

Remaining heat $Q = 184000 - 16000 . 56$

$= 167999 . 44J$

Heat required to melt ice completely at $0 ° C$,

${Q}_{2} = 0 . 600 \times 336000 = 201600J$ needed which is more than $Q$, so $100$ ice is not melted.

For amount of melted ice, we can write

$167999 . 44 = m \times 336000 ⇒ m = 0 . 4999kg$

$∴$ mass of water $= 0 . 4999kg$

Mass of ice remaining $= 0 . 1001$

$∴$ Ratio $= \dfrac{0 . 1001}{0 . 4999} \approx 1: 5$