Respuesta :
Answer:
[tex]Q=35011.6\ J[/tex]
Explanation:
Given:
- mass of ice, [tex]m=50\ g[/tex]
- initial temperature of ice, [tex]T_i=-30^{\circ}C[/tex]
- final temperature of liquid water, [tex]T_f=73^{\circ}C[/tex]
- heat of fusion of ice, [tex]L=333\ J.g^{-1}[/tex]
- specific heat capacity of ice, [tex]c_i=2.06\ J.g^{-1}[/tex]
- specific heat capacity of liquid water, [tex]c_w=4.184\ J.g^{-1}.K^{-1}[/tex]
Now, total heat energy required get to the final state:
[tex]Q=m(c_i.\Delta T_i+L+c_w.\Delta T_w)[/tex]
where:
[tex]\Delta T_w=[/tex] change in temperature of water from 0 to 73 degree C
[tex]\Delta T_i=[/tex] change in temperature of ice from -30 to 0 degree C
[tex]\therefore Q=50(2.06\times 30+333+4.184\times 73)[/tex]
[tex]Q=35011.6\ J[/tex]
Answer:
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
