Given:
ΔQ=184×103J,
m=0.600kgat−12∘C,
S=2222.3J{\mathrm{kg}}^{–1}^{\circ}{C}^{–1} and
L=336×103Jkg−1
Heat required to convert ice from –12∘C to ice at 0∘C will be,
Q1=0.600×2222.3×12=16000.56J
Remaining heat Q=184000−16000.56
=167999.44J
Heat required to melt ice completely at 0∘C,
Q2=0.600×336000=201600J needed which is more than Q, so 100 ice is not melted.
For amount of melted ice, we can write
167999.44=m×336000⇒m=0.4999kg
∴ mass of water =0.4999kg
Mass of ice remaining =0.1001
∴ Ratio =0.49990.1001≈1:5