Question

(a) How much heat transfer is necessary to raise the temperature of a 0.200-kg piece of ice from −20.0ºC to 130ºC, including the energy needed for phase changes? (b) How much time is required for each stage, assuming a constant 20.0 kJ/s rate of heat transfer?

Answer 1

For a: The heat required for the given process is 623.74 kJ

For b: The time required for each process is calculated below.

Step-by-step explanation:

• For a:

The processes involved in the given problem are:

`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)`

Pressure is taken as constant.

To calculate the amount of heat absorbed at different temperature, we use the equation:

`q=m* C_(p,m)* (T_(2)-T_(1))` .......(1)

where,

q = amount of heat absorbed = ?

`C_(p,m)` = specific heat capacity of medium

m = mass of water/ice

`T_2` = final temperature

`T_1` = initial temperature

To calculate the amount of heat released at same temperature, we use the equation:

`q=m* L_(f,v)` ......(2)

where,

q = amount of heat absorbed = ?

m = mass of water/ice

`L_(f,v)` = 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:

`m=0.200kg=200g\\C_(p,s)=2.03J/g^oC\\T_1=-20^oC\\T_2=0^oC`

Putting values in equation 1, we get:

`q_1=200* 2.03J/g^oC* (0-(-20))^oC\\\\q_1=8120J=8.120kJ`

• For process 2:

We are given:

`m=200g\\L_f=334J/g`

Putting values in equation 2, we get:

`q_2=200g* 334J/g=66800J=66.800kJ`

• For process 3:

We are given:

`m=200g\\C_(p,l)=4.184J/g^oC\\T_1=0^oC\\T_2=100^oC`

Putting values in equation 1, we get:

`q_3=200g* 4.184J/g^oC* (100-(0))^oC\\\\q_3=83680J=83.860kJ`

• For process 4:

We are given:

`m=200g\\L_v=2266J/g`

Putting values in equation 2, we get:

`q_4=200g* 2266J/g=453200J=453.200kJ`

• For process 5:

We are given:

`m=200g\\C_(p,g)=1.99J/g^oC\\T_1=100^oC\\T_2=130^oC`

Putting values in equation 1, we get:

`q_5=200g* 1.99J/g^oC* (130-(100))^oC\\\\q_5=11940J=11.940kJ`

Total heat absorbed =
`q_1+q_2+q_3+q_4+q_5`

Total heat absorbed =
`[8.120+66.800+83.680+453.200+11.940]kJ=623.74kJ`

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 =
`(1s)/(20kJ)* 8.120kJ=0.406s`

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 =
`(1s)/(20kJ)* 66.800kJ=3.34s`

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 =
`(1s)/(20kJ)* 83.680kJ=4.184s`

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 =
`(1s)/(20kJ)* 453.200kJ=22.66s`

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 =
`(1s)/(20kJ)* 11.940kJ=0.597s`

Time required for heat transfer of process 5 is 0.597 seconds

Answer 2

a)Q=167.5 KJ

b)t=8.375 s

Step-by-step 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