Respuesta :
Answer:
a)Q=167.5 KJ
b)t=8.375 s
Explanation:
a)Given that
mass m= 0.2 kg
We know that
Cp for ice is 2.1 KJ/Kg.K
Cp for water is 4.718 KJ/Kg.K
Cp for steam is 1.9 KJ/Kg.K
Latent heat of ice=336 KJ/kg
First ice will take sensible heat and increases temperature from −20.0ºC to 0ºC after that it will transfer phase from solid to liquid at constant temperature and will take latent heat after that water will take sensible heat and increases temperature from 0ºC to 100ºC. Above 100ºC water will convert in to vapor and vapor will take sensible heat increases temperature from 100ºC to 130ºC .
Sensible heat = m CpΔT KJ
Latent heat = m L.H. KJ
So total heat
Q=0.2(2.1 x 20 + 336 + 4.178 x 100 + 1.9 x 130) KJ
Q=167.5 KJ
b)
Heat transfer = 20 KJ/s
Lets time taken is t
20 t =167.5
t=8.375 s
Answer:
For a: The heat required for the given process is 623.74 kJ
For b: The time required for each process is calculated below.
Explanation:
- For a:
The processes involved in the given problem are:
[tex]1.)H_2O(s)(-20^oC)\rightarrow H_2O(s)(0^oC)\\2.)H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\3.)H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)\\4.)H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)\\5.)H_2O(g)(100^oC)\rightarrow H_2O(g)(130^oC)[/tex]
Pressure is taken as constant.
To calculate the amount of heat absorbed at different temperature, we use the equation:
[tex]q=m\times C_{p,m}\times (T_{2}-T_{1})[/tex] .......(1)
where,
q = amount of heat absorbed = ?
[tex]C_{p,m}[/tex] = specific heat capacity of medium
m = mass of water/ice
[tex]T_2[/tex] = final temperature
[tex]T_1[/tex] = initial temperature
To calculate the amount of heat released at same temperature, we use the equation:
[tex]q=m\times L_{f,v}[/tex] ......(2)
where,
q = amount of heat absorbed = ?
m = mass of water/ice
[tex]L_{f,v}[/tex] = latent heat of fusion or vaporization
Calculating the heat absorbed for each process:
- For process 1:
Conversion factor used: 1 kg = 1000 g
1 kJ = 1000 J
We are given:
[tex]m=0.200kg=200g\\C_{p,s}=2.03J/g^oC\\T_1=-20^oC\\T_2=0^oC[/tex]
Putting values in equation 1, we get:
[tex]q_1=200\times 2.03J/g^oC\times (0-(-20))^oC\\\\q_1=8120J=8.120kJ[/tex]
- For process 2:
We are given:
[tex]m=200g\\L_f=334J/g[/tex]
Putting values in equation 2, we get:
[tex]q_2=200g\times 334J/g=66800J=66.800kJ[/tex]
- For process 3:
We are given:
[tex]m=200g\\C_{p,l}=4.184J/g^oC\\T_1=0^oC\\T_2=100^oC[/tex]
Putting values in equation 1, we get:
[tex]q_3=200g\times 4.184J/g^oC\times (100-(0))^oC\\\\q_3=83680J=83.860kJ[/tex]
- For process 4:
We are given:
[tex]m=200g\\L_v=2266J/g[/tex]
Putting values in equation 2, we get:
[tex]q_4=200g\times 2266J/g=453200J=453.200kJ[/tex]
- For process 5:
We are given:
[tex]m=200g\\C_{p,g}=1.99J/g^oC\\T_1=100^oC\\T_2=130^oC[/tex]
Putting values in equation 1, we get:
[tex]q_5=200g\times 1.99J/g^oC\times (130-(100))^oC\\\\q_5=11940J=11.940kJ[/tex]
Total heat absorbed = [tex]q_1+q_2+q_3+q_4+q_5[/tex]
Total heat absorbed = [tex][8.120+66.800+83.680+453.200+11.940]kJ=623.74kJ[/tex]
Hence, the heat required for the given process is 623.74 kJ
- For b:
We are given:
Rate of heat transfer = 20.0 kJ/s
To calculate the time required for each stage, we apply unitary method:
- For process 1:
Heat transferred = 8.120 kJ
When 20 kJ of heat is transferred in 1 sec
So, 8.120 kJ of heat will be transferred in = [tex]\frac{1s}{20kJ}\times 8.120kJ=0.406s[/tex]
Time required for heat transfer of process 1 is 0.406 seconds
- For process 2:
Heat transferred = 66.800 kJ
When 20 kJ of heat is transferred in 1 sec
So, 66.800 kJ of heat will be transferred in = [tex]\frac{1s}{20kJ}\times 66.800kJ=3.34s[/tex]
Time required for heat transfer of process 2 is 3.34 seconds
- For process 3:
Heat transferred = 83.680 kJ
When 20 kJ of heat is transferred in 1 sec
So, 83.680 kJ of heat will be transferred in = [tex]\frac{1s}{20kJ}\times 83.680kJ=4.184s[/tex]
Time required for heat transfer of process 3 is 4.184 seconds
- For process 4:
Heat transferred = 453.200 kJ
When 20 kJ of heat is transferred in 1 sec
So, 453.200 kJ of heat will be transferred in = [tex]\frac{1s}{20kJ}\times 453.200kJ=22.66s[/tex]
Time required for heat transfer of process 4 is 22.66 seconds
- For process 5:
Heat transferred = 11.940 kJ
When 20 kJ of heat is transferred in 1 sec
So, 11.940 kJ of heat will be transferred in = [tex]\frac{1s}{20kJ}\times 11.940kJ=0.597s[/tex]
Time required for heat transfer of process 5 is 0.597 seconds
