How much energy is required to heat a chunk of ice that weighs 510g from -256°c to 195°c

The ice undergoes several transformations. First it absorbs energy and its temperature to raise the temperature to 0°C the melting point of water. Then the ice absorbs latent heat of fusion to melt, then absorbs heat to raise the temperature to 100°C the boiling point of water. then the steam is heated to 195°C

Therefore the following expression is used to calculate the heat required to make these changes

MC∅(ice)+MLf(ice)+MC∅(water)+MLv(water)+MC∅(water vapor)

0.51kg×2.108kJ/KgK×(0--256)K+0.51×333.55kJ/Kg+0.51kg×4.2 kJ/KgK×(100-0)+0.51kg×2260kJ/kg+ 0.51kg×1.996kJ/kgK× (195-100)K

= 1908.84kJ

a

explanation:

warm front forms when a moist, warm air mass slides up and over a cold air mass. as the warm air mass rises, it condenses into a broad area of clouds. a warm front brings gentle rain or light snow, followed by warmer, milder weather.

hope i !