Part A: Find the amount of heat that must be extracted from 1.5 kg of steam at 130 ∘C to convert it to ice at 0.0 ∘C.
Part B: What speed would this 1.5-kg block of ice have if its translational kinetic energy were equal to the thermal energy calculated in part A?
TRUST THE PROCESS
a.) (this is the process you'll use to find the heat needed and the heat extracted)
1.ice at -20∘C → heated to ice at 0∘C
2.ice at 0∘C → heated to water at 0∘C
3.water at 0∘C → heated to water at 100∘C
4.water at 100∘C → heated to vapor (steam) at 100∘C
5.vapor (steam) at 100∘C → heated to vapor at 130∘C
1. m=1.5 Q=m(c)(delta T)
c(specific heat of ice)= 2090 =(1.5)(2090)(20∘C)
delta T = 0-(-20)= 20∘C =62700 J
2. m=1.5 Q=m(Lf)
Lf(fusion)= 33.5*10^4 =(1.5)(33.5*10^4)
=502500 J
3. m=1.5 Q=m(c)(delta T)
c(specific heat of water)= 4186 =(1.5)(4186)(100)
delta T= 100- 0= 100∘C =627900 J
4. m=1.5 Q=m(Lv)
Lv(vaporization)= 22.6*10^5 =(1.5)(22.6*10^5)
=3390000 J
5. m=1.5 Q=m(c)(delta T)
c(specific heat of vapor)=2010 =(1.5)(2010)(30)
delta T=130-100= 30∘C =90450 J
Now you are going to add all the J's up
62700 J + 502500 J + 627900 J + 3390000 J + 90450 J
Q =4673550 J
There ya go hope that helps and sorry not sure how to do part b...
kinda posted funky but hope that helped a lil
To find the amount of heat that must be extracted from the steam to convert it to ice, we need to calculate the heat required for two processes: cooling the steam to its condensation temperature and then converting it from steam to ice.
Part A:
1. Cooling the steam to its condensation temperature:
First, we need to calculate the amount of heat required to lower the temperature of the steam from 130°C to 100°C using the specific heat capacity of steam.
The specific heat capacity of water vapor (steam) is approximately 2,080 J/(kg⋅°C).
The formula for heat transfer is Q = mcΔT, where Q is the heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
Q1 = (mass of steam) * (specific heat capacity of steam) * (change in temperature)
= 1.5 kg * 2080 J/(kg⋅°C) * (100°C - 130°C)
= 1.5 kg * 2080 J/(kg⋅°C) * (-30°C)
= -93,600 J
Note: The negative sign indicates that heat is being extracted from the steam.
2. Converting the steam to ice:
To convert the steam to ice, we need to calculate the heat of vaporization required. The heat of vaporization (or latent heat) of water is approximately 2,260,000 J/kg.
Q2 = (mass of steam) * (heat of vaporization of water)
= 1.5 kg * 2,260,000 J/kg
= 3,390,000 J
Therefore, the total heat that must be extracted from the steam is:
Q_total = Q1 + Q2
= -93,600 J + 3,390,000 J
= 3,296,400 J
Therefore, approximately 3,296,400 J of heat must be extracted from 1.5 kg of steam at 130°C to convert it to ice at 0°C.
Part B:
To find the speed of the block of ice that would have translational kinetic energy equal to the thermal energy calculated in Part A, we can use the equation:
Kinetic energy (KE) = 1/2 * mass * velocity^2
Since the mass is given as 1.5 kg and we want the KE to be equal to 3,296,400 J, we can solve for velocity:
3,296,400 J = 1/2 * 1.5 kg * velocity^2
Rearranging the equation:
velocity^2 = (2 * 3,296,400 J) / (1.5 kg)
velocity^2 = 4,394,933.33 m^2/s^2
Taking the square root of both sides to find the velocity:
velocity = sqrt(4,394,933.33 m^2/s^2)
≈ 2,095.8 m/s
Therefore, the speed of the 1.5 kg block of ice would be approximately 2,095.8 m/s if its translational kinetic energy were equal to the thermal energy calculated in Part A.