How Much Heat Must Be Removed from 456 G $25.0 ^ { \circ } \mathrm { C }$

How much heat must be removed from 456 g of water at $25.0 ^ { \circ } \mathrm { C }$ to change it into ice at $- 10.0 ^ { \circ } \mathrm { C } ?$ The specific heat of ice is $2090 \mathrm {~J} / \mathrm { kg } \cdot \mathrm { K }$ the latent heat of fusion of water is $33.5 \times 10 ^ { 4 } \mathrm {~J} / \mathrm { kg }$ and the specific heat of water is $4186 \mathrm {~J} / \mathrm { kg } \cdot \mathrm { K }$

A) 153 kJ
B) 210 kJ
C) 57.3 kJ
D) 47.7 kJ
E) 105 kJ