Question

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

Answer 1

`Q=35011.6\ J`

Step-by-step explanation:

Given:

• mass of ice,
  `m=50\ g`

• initial temperature of ice,
  `T_i=-30^(\circ)C`

• final temperature of liquid water,
  `T_f=73^(\circ)C`

• heat of fusion of ice,
  `L=333\ J.g^(-1)`

• specific heat capacity of ice,
  `c_i=2.06\ J.g^(-1)`

• specific heat capacity of liquid water,
  `c_w=4.184\ J.g^(-1).K^(-1)`

Now, total heat energy required get to the final state:

`Q=m(c_i.\Delta T_i+L+c_w.\Delta T_w)`

where:

`\Delta T_w=` change in temperature of water from 0 to 73 degree C

`\Delta T_i=` change in temperature of ice from -30 to 0 degree C

`\therefore Q=50(2.06* 30+333+4.184* 73)`

`Q=35011.6\ J`

Answer 2

The heat energy required is 350.1J

Step-by-step 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