Calculate the energy that is required to change 50.0 g ice at -30.0°c to a liquid at 73.0°c. the heat of fusion = 333 j/g, the heat of vaporization = 2256 j/g, and the specific heat capacities of ice = 2.06 j/gk and liquid water = 4.184 j/gk

The heat energy required is 350.1J

Explanation:

Given data

Mass of ice =50g---kg =50/1000= 0.05kg

Temperature of ice T1= - 30°c

Temperature of ice T2=0°c

Temperature of liquid T3=0°c

Temperature of water T4= 73°c

The heat of fusion = 333 J/g

heat of vaporization = 2256 J/g, and the specific heat capacities of ice = 2.06 J/gK

and liquid water = 4.184 J/gK

It will be a good idea to first understand the path this process will follow

Ice at - 30°c - - ice at 0°c

Ice at 0°c - - - - liquid at 0°c fusion

Liquid at 0°c -- - liquid at 73°c

First, you have to calculate the heat absorbed by ice going from -30 C to 0 C. Use the equation:

q = m c (T2-T1)

q= 0.5*2.06(0-(-30))

q= 30.9J

Then, calculate the heat required to melt that ice at 0C. Use the equation:

q = m *(heat of fusion)

q=0.5*333

q=166.5J

Then, calculate the heat required to raise the temperature of water from 0°C to 73°C. Use

q = m c (T4-T3)

q= 0.5*4.184(73-0)

q= 152.7J

Finally, we will sum up the heat required

Total heat energy required = 30.9+166.5+152.7= 350.1J

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