When you get out of a pool, even the slightest breeze will make you feel cold. Why is this?

Answer: Evaporation of water is an endothermic process, meaning it sucks in heat from the environment, which includes your skin which is at a higher temperature than the water. Since it sucks heat energy form your skin, you feel cold.

Next Question: (From Shortley and Williams Elements of Physics, 3rd Edition) A layer of ice on a pond is 2 cm thick. When the upper surface of the ice is at -15C and the temperature of the water just below the ice is 0C, at what rate does the ice become thicker? Express your answer in units of [cm/min.]. Show your work.

[math]\frac{dQ}{dt} = k\frac{A(T_h - T_c)}{L}[/math]
Plugging in the numbers given and knowing that the thermal conductivity of ice is approximately 2.25 W/mk:
[math]\frac{dQ}{dt} = 2.25\frac{0.0001(0 - (-15))}{0.02} = 0.16875 W[/math]
These means that 0.16875 Joules are being transferred across each cm^2 of ice every second. We know that the the latent heat of fusion is 334 Joules/gram and that the density of ice is 0.92 g/mL. Now we can find the mass and volume of ice formed:
[math]\frac{dM}{dt} = \frac{dQ}{dt} * \frac{1}{L_{fusion}} = 0.000505 g/s[/math]
[math]\frac{dV}{dt} = \frac{dM}{dt} * \frac{1}{\rho} = 0.000549 cm^3/s[/math]
The area is 1 cm^2 so we can plug that into the next equation. To find the rate of ice formation:
[math]\frac{dL}{dt} = \frac{dV}{dt} * \frac{1}{dA} = 0.000549 cm/s[/math]
This answer is in centimeters per second, so multiply by 60 to find the number of centimeters of ice formed per minute:
[b]0.03295 cm/min[/b]

"The fault, dear Brutus, is not in our stars,
But in ourselves, that we are underlings."

University of Texas at Austin '23
Seven Lakes High School '19

When you get out of a pool, even the slightest breeze will make you feel cold. Why is this?

Answer: Evaporation of water is an endothermic process, meaning it sucks in heat from the environment, which includes your skin which is at a higher temperature than the water. Since it sucks heat energy form your skin, you feel cold.

Next Question: (From Shortley and Williams Elements of Physics, 3rd Edition) A layer of ice on a pond is 2 cm thick. When the upper surface of the ice is at -15C and the temperature of the water just below the ice is 0C, at what rate does the ice become thicker? Express your answer in units of [cm/min.]. Show your work.

[math]\frac{dQ}{dt} = k\frac{A(T_h - T_c)}{L}[/math]
Plugging in the numbers given and knowing that the thermal conductivity of ice is approximately 2.25 W/mk:
[math]\frac{dQ}{dt} = 2.25\frac{0.0001(0 - (-15))}{0.02} = 0.16875 W[/math]
These means that 0.16875 Joules are being transferred across each cm^2 of ice every second. We know that the the latent heat of fusion is 334 Joules/gram and that the density of ice is 0.92 g/mL. Now we can find the mass and volume of ice formed:
[math]\frac{dM}{dt} = \frac{dQ}{dt} * \frac{1}{L_{fusion}} = 0.000505 g/s[/math]
[math]\frac{dV}{dt} = \frac{dM}{dt} * \frac{1}{\rho} = 0.000549 cm^3/s[/math]
The area is 1 cm^2 so we can plug that into the next equation. To find the rate of ice formation:
[math]\frac{dL}{dt} = \frac{dV}{dt} * \frac{1}{dA} = 0.000549 cm/s[/math]
This answer is in centimeters per second, so multiply by 60 to find the number of centimeters of ice formed per minute:
[b]0.03295 cm/min[/b]

You have a piston filled with 8 moles of a monatomic ideal gas at a temperature of 278 K. It is completely insulated (meaning that there is no heat transfer between the walls of the piston) and it is at rest in a vacuum. The piston has a diameter of 25 cm and has a mass of 4.5 kg. You then apply 50 N of force to the top of the piston.

Determine the change in temperature of the gas.

"The fault, dear Brutus, is not in our stars,
But in ourselves, that we are underlings."

University of Texas at Austin '23
Seven Lakes High School '19

You have a piston filled with 8 moles of a monatomic ideal gas at a temperature of 278 K. It is completely insulated (meaning that there is no heat transfer between the walls of the piston) and it is at rest in a vacuum. The piston has a diameter of 25 cm and has a mass of 4.5 kg. You then apply 50 N of force to the top of the piston.

Determine the change in temperature of the gas.

P = F/A = (4.5 kg * 9.8 m/s^2)/(0.0625pi m^2) = 224.599 Pa
V = nRT/P = (8 mol)R(278 K)/224.599 Pa =
82.331 m^3
P2 = F/A = (4.5 kg * 9.8 m/s^2 + 50 N)/(0.0625pi m^2) = 479.247 Pa
Since it's adiabatic compression on a monatomic ideal gas, PV^(5/3) = (P2)(V2)^(5/3);
V2 = (PV^(5/3)/P2)^(3/5) = 52.248 m^3
PV/T = (P2)(V2)/(T2); T2 = (P2)(V2)T/(PV)
= 376.448 K
T2 - T = [b]98 K[/b]

You have a piston filled with 8 moles of a monatomic ideal gas at a temperature of 278 K. It is completely insulated (meaning that there is no heat transfer between the walls of the piston) and it is at rest in a vacuum. The piston has a diameter of 25 cm and has a mass of 4.5 kg. You then apply 50 N of force to the top of the piston.

Determine the change in temperature of the gas.

P = F/A = (4.5 kg * 9.8 m/s^2)/(0.0625pi m^2) = 224.599 Pa
V = nRT/P = (8 mol)R(278 K)/224.599 Pa =
82.331 m^3
P2 = F/A = (4.5 kg * 9.8 m/s^2 + 50 N)/(0.0625pi m^2) = 479.247 Pa
Since it's adiabatic compression on a monatomic ideal gas, PV^(5/3) = (P2)(V2)^(5/3);
V2 = (PV^(5/3)/P2)^(3/5) = 52.248 m^3
PV/T = (P2)(V2)/(T2); T2 = (P2)(V2)T/(PV)
= 376.448 K
T2 - T = [b]98 K[/b]

You got it right, nice job!

One thing I noticed while looking through your work is that when you were calculating for pressure and area, you used the diameter instead of the radius of the piston. This didn't really matter in the end since the pressures got divided out in the final formula (causing the areas to cancel out), but it's something that you should definitely watch for.

"The fault, dear Brutus, is not in our stars,
But in ourselves, that we are underlings."

University of Texas at Austin '23
Seven Lakes High School '19

This may be more than anything I personally remember ever seeing on a thermo test, but yet still potentially a feasible test question (for C division at least), and I have recently developed an interest in boilers.

A sealed, perfectly insulated boiler is a cylinder with the (interior) dimensions of 1m in radius, 5m height and is 70% full of water at 25C. Assume ambient pressure of 1atm and ignore the heat capacity of the boiler. A furnace burns methane at a rate of 15MW and transfers 60% of the heat to the boiler.

a. What is the minimum stoichiometric air flow to the furnace in m^3/minute?

b. How fast will the furnace initially heat in degrees Celsius per minute? When will it reach boiling point?

c. Describe what happens inside the boiler as heat continues to be added.

d. Once the boiler reaches a pressure of 3MPa (absolute), a valve is kept opened such to maintain a constant pressure. Once most of the air is removed from the boiler, determine the following: the temperature of the boiler, the outflow rate of steam in kg/minute, an estimate of the time before the boiler runs dry (you may ignore factors like changing steam volume and water evaporated to reach steady state).

e. Feed water is now fed into the boiler at 20C and an equilibrium state with a water level of 50% is reached. Determine the new steam outflow rate.

f. What else can be done for the boiler to produce more steam without increasing the furnace output?

g. (Extra) The boiler fails catastrophically under full operating conditions. Estimate the size of the resulting steam cloud, neglecting the work done on the (former) boiler and the immediate surroundings.

a. The equation for methane combustion is [math]CH_4+2O_2 \to CO_2+2H_2O[/math]. The enthalpy of combustion of methane is [math]822kJ/mol_{CH_4}[/math]. [math]15MW/(-822kJ/mol_{CH_4})*2mol_{O_2}/mol_{CH_4}=2200 mol_{O_2}/minute[/math]. If gas at standard conditions is [math]22.4 L/mol[/math] and air is 21% oxygen, we get that there is [math]110L_{air}/mol_{O_2}[/math].
[math]2200mol_{O_2}/minute*(110L_{air}/mol_{O_2})=230 m^3_{air}/minute[/math]
b. The volume of the boiler is [math]pi*(1m)^2*5m=15.7m^3[/math] (will assume measurements had 2-3 sig figs if I wrote more carefully). At standard density, that gives a mass of water of [math]15.7m^3*70% water*1000kg/m^3=11100kg water[/math]. The water receives [math]15MW*60%=9MW[/math] heat.
[math]9MW/(4.186 J/(kg \degree C))/11100kg=11.7 \degree C/minute[/math]. The difference in temperature is [math]100 \degree C-25 \degree C=75 \degree C[/math]. [math]75 \degree C/(11.7 \degree C/minute)=384 seconds[/math].
c. Steam evaporates until the pressure inside the boiler equals the vapor pressure of water at the current temperature. As heat continues to be added to the boiler, this pressure continues to increase.
d. This is where you can pull out the standard enthalpies from the steam tables (find a good one or two, learn how to read it, and add it your binder in case). According to [url=http://www.steamshed.com/pdf/034SteamTablesAsme.pdf]Page 19 of this one[/url], at 3.0MPa, saturated steam has a temperature of [math]233 \degree C[/math] and a standard enthalpy of [math]2803 kJ/kg[/math], whereas the water has a standard enthalpy of [math]1008 kJ/kg[/math].
For every kilogram of steam released, at least a kilogram of water must evaporate to take its place, so the furnace must match the difference in enthalpy, which is [math]2803kJ/kg- 1008kJ/kg=1795kJ/kg[/math]. [math]9MW/(1795kJ/kg)=301kg/minute[/math]. As a first order approximation, 11100kg will last [math]11100kg/(301kg/minute)=37 minutes[/math]. If you wish to be more precise, as the last liquid water vaporizes, you still have [math]15.7m^3[/math] of pressurized steam when the valve would have to close to maintain pressure. Also according to the steam table, the steam takes up [math]0.0667m^3/kg[/math], so [math]15.7m^3/(0.0667m^3/kg)=235kg[/math] of water is still in the boiler as steam.
e. Now instead of replacing the change in enthalpy from [math]233\degree C[/math] water into saturated steam, you additionally have to heat the replacement water up to that temperature from [math]20\degree C[/math], which has a standard enthalpy of [math]83.9 kJ/kg[/math] (correct me if I am significantly wrong to use the one at low pressure). Thus the total replacement enthalpy per kg of steam is [math]2803kJ/kg- 83.9kJ/kg=2719kJ/kg[/math]. [math]9MW/(2719kJ/kg)=199kg/minute[/math]
f. The intended answer was to preheat the feed water by running the pipe through the furnace exhaust, as there is excess heat and you are losing steam production to heat the water inside the boiler.
Not attempting g today.

Last edited by Schrodingerscat on January 16th, 2018, 9:49 pm, edited 1 time in total.

This may be more than anything I personally remember ever seeing on a thermo test, but yet still potentially a feasible test question (for C division at least), and I have recently developed an interest in boilers.

A sealed, perfectly insulated boiler is a cylinder with the (interior) dimensions of 1m in radius, 5m height and is 70% full of water at 25C. Assume ambient pressure of 1atm and ignore the heat capacity of the boiler. A furnace burns methane at a rate of 15MW and transfers 60% of the heat to the boiler.

a. What is the minimum stoichiometric air flow to the furnace in m^3/minute?

b. How fast will the furnace initially heat in degrees Celsius per minute? When will it reach boiling point?

c. Describe what happens inside the boiler as heat continues to be added.

d. Once the boiler reaches a pressure of 3MPa (absolute), a valve is kept opened such to maintain a constant pressure. Once most of the air is removed from the boiler, determine the following: the temperature of the boiler, the outflow rate of steam in kg/minute, an estimate of the time before the boiler runs dry (you may ignore factors like changing steam volume and water evaporated to reach steady state).

e. Feed water is now fed into the boiler at 20C and an equilibrium state with a water level of 50% is reached. Determine the new steam outflow rate.

f. What else can be done for the boiler to produce more steam without increasing the furnace output?

g. (Extra) The boiler fails catastrophically under full operating conditions. Estimate the size of the resulting steam cloud, neglecting the work done on the (former) boiler and the immediate surroundings.

a. Looking up the combustion equation of methane, it's CH4 + 2O2 -> CO2 + 2H2O + 882 kJ. The relevant part of the equation is 2O2 -> 882 kJ. Thus, 441 kJ/mol O. 15 MW = 15E3 kJ/s = 15E3 * 60 kJ/min = 9E5 kJ/min. 9E5 kJ/min / 441 kJ/mol = 2041 mol/min. PV = nRT. I don't know the temperature of the oxygen gas, so I'm stuck here. :?
b. 15 MW * 0.6 = 15E6 W * 0.6 = 9E6 J/s = 9E6 * 60 J/min. 9E6 * 3600 J/min / (4184 J/kg*oC) = 9E6 * 3600 / 4184 kg*oC/min = 7.74E6 kg*oC/min. The mass of the water can be found using the volume of the water, V = .7 * 5pi m^3 = 11.0 m^3. m = 11.0 m^3 * 1000 kg/m^3 = 11000 kg. (7.74E6 kg*oC/min) / (11000 kg) = [b]704 oC/min[/b] (I don't know how realistic that is.) 75 oC / (704 oC/min) = .107 min = [b]6.39 seconds[/b] (Again, I don't know how realistic that is.)
c. The steam mixes with the air already inside the boiler, and as the gases continue to get heated, their particles start getting faster and faster. The pressure and temperature of the mixture goes up while the volume remains constant.