A tankless water heater heats water as it flows through the device. The
water circulates through a copper heat exchanger. For a particular water heater
the flow rate is 8.7 liter per minute. The temperature rise of the water is 25
^{o}C. The water heater uses gas to heat the water in the copper coils
and is 80% efficient. What is the gas energy input rate in units of kW?

conversion: 1 kcal = 4186 J

Solution:

- Concepts:

Energy conservation, specific heat c - Reasoning:

Water flow rate (kg/s) * c(J/(kg^{o}C) * temperature change (^{o}C) = power input (W) * efficiency. - Details of the calculation:

Water flow rate: 8.7 kg/minute = (8.7/60) kg/s,

specific heat of water: c = 4186 J/(kg^{o}C).

P = ((8.7/60) kg/s)*(4186 J/(kg^{o}C))*(25^{o}C)/0.8 = 18967 W ≈ 19 kW.

Particles evaporating in the z-direction from a hole in a container with an
ideal gas at temperature T.

(a) Compute the speed distribution, f*(v), of the evaporating particles.

(b) Compute the mean z-component of the velocity, <v_{z}>* of the
escaping particles.

(c) Compute the mean energy of the particles leaving the container.

Assume that the equilibrium in the interior of the container is not disturbed by
the evaporating particles.

Maxwell-Boltzmann speed distribution: f(v) = (m/(2πkT))^{3/2 }4πv^{2}exp(-mv^{2}/(2kT)

Solution:

- Concepts:

The Maxwell-Boltzmann speed distribution, geometry - Reasoning:

The atoms escape through a small hole of area A. We need to find the rate at which atoms with different v or v_{z}escape from the hole. - Details of the calculation:

Consider an area dA and gas atoms with speeds between v and v + dv in a ring at a distance between r and r + dr from dA. Assume that the gas atoms travel in straight lines. The probability that gas atoms from the ring will reach the area dA is dA cosθ/(4πr^{2}). This is the solid angle subtended by dA at any dV in the ring.

The number of atoms from the ring reaching dA with speeds between v and v + dv is

nf(v)dv2πr^{2}sinθ dr dθ dA cosθ/(4πr^{2}) = ½nf(v)dv sinθ dr dθ dA cosθ.

Here n = N/V is the density of the gas.

The total number of atoms with speeds between v and v + dv reaching dA in a time interval dt comes from rings with radii between r = 0 and dr = vdt. Per unit area per unit time it is therefore given by

∫_{0}^{π/2}dθ nf(v)dv v sinθ cosθ/2 = ¼ n v f(v) dv.

The escape rate per unit area per unit time is

R_{esc}= ½n∫_{0}^{∞}dv ∫_{0}^{π/2}dθ sinθ cosθ v f(v) = ¼ n ∫_{0}^{∞}vf(v) dv = n (kT/(2mπ))^{½}.

We therefore have f*(v) = ¼ n v f(v)/R_{esc}= ½(m^{2}/(kT)^{2})^{ }v^{3}exp(-mv^{2}/(2kT).

(b) Atoms from a ring reaching dA with speed v have a velocity component v_{z}= vcosθ.

The average velocity component <v_{z}> for the particles escaping from the hole is given by

[½n∫_{0}^{∞}dv ∫_{0}^{π/2}dθ sinθ cos^{2}θ v^{2}f(v)]/R_{esc}

= (3/4)(2πkT/m)^{½}[∫_{0}^{π/2}dθ sinθ cos^{2}θ]/[ ∫_{0}^{π/2}dθ sinθ cosθ] = ½(2πkT/m)^{½}.

(c) The mean energy of the particles is given by <E> = ½m∫_{0}^{∞}v^{2}f*(v) dv.

<E> = ¼(m^{3}/(kT)^{2})∫_{0}^{∞}v^{5}exp(-mv^{2}/(2kT) dv = 2kT.

A small air bubble of initial radius r = 1 cm is introduced at the bottom of
a lake that is 20 m deep. The bubble expands as it rises slowly. Assume the lake
has the same temperature everywhere.

(a) What is the pressure at the bottom of the lake?

(b) What is the radius of the bubble after it rises to the surface?

Solution:

- Concepts:

Pressure in a fluid, the ideal gas law - Reasoning:

The pressure at the bottom of a liquid column is P_{bottom}= P_{top}+ ρgh. - Details of the calculation:

P_{bottom}= P_{surf}+ ρgz = 1 atm + (1000 kg/m^{3})(9.8 m/s^{2})(20 m) = 1 atm + 196000 Pa.

1 atm = 1.013 * 10^{5}Pa.

P_{bottom}= 2.93 atm.

Ideal gas law: PV = NkT = constant.

P_{i}V_{i}= P_{f}V_{f}.

V_{f}= P_{i}/P_{f}V_{i}= (4/3)πr_{f}^{3}= [(2.93 atm)/(1 atm)]* (4/3) π(0.01)^{3}.

r_{f}= 1.43 cm.

Estimate (roughly) the net rate of heat loss by your body due to radiation
when you are in a room with a temperature of 10 ^{o}C.

Solution:

- Concepts:

Stefan-Boltzmann law - Reasoning:

Heat loss = σ* T^{4}_{body}* Area, heat gain = σ* T^{4}_{surrounding}* Area.

Net loss = σ* (T^{4}_{body }- T^{4}_{surrounding})*Area. - Details of the calculation:

T_{body}~ 33^{o}C_{ }= 306 K (skin temperature).

Area ~ 1.75 m^{2}(skin surface area).

Net loss ~ 5.67*10^{-8}(306^{4}- 283^{4})*1.75 W = 234 W.

During phase transition, the change in the pressure P and temperature T can be expressed by the Clausius-Clapeyron relation,

dP/dT = L/(T ∆V),

where L is the latent heat and ∆V is the change in volume.

(a)
How much pressure does one have to put on an ice cube to make it melt at -1^{o} C?

The density of ice is 917 kg/m^{3}, and the latent heat of 1 kg of
ice melting is 333000 J.

(b) Approximately how deep under a glacier does it have to be before the
weight of the ice above gives the pressure you found in part (a)?

Solution:

- Concepts:

The Clausius-Clapeyron relation - Reasoning:

For small changes of the melting point temperature we expand ∆P = (dP/dT)∆T. - Details of the calculation:

(a) The density of water is 1000 kg/m^{3}. The volume of 1 kg of water is 1/1000 m^{3}.

From the density of ice 917 kg/m^{3}, the volume of 1 kg of ice is 1/917 m^{3}.

For phase transition of 1 kg of ice to water ∆V =(1/1000 - 1/917) m^{3}= -9.051*10^{-5}m^{3}.

dP/dT = (333000 J)/(273 K * ∆V) = -1.35*10^{7}Pa/K = -135 bar.

For the melting point decrease by 1^{o}C we have ∆P = (dP/dT)∆T = 135 bar.

For the melting point decrease by 1^{o}C, the pressure should increase by 135 bar.

(b) To reduce the melting point by 1^{o}C, the pressure under a glacier should be ~136 bar.

P_{under_glacier}= P_{air }+ P_{glacier}= 1 bar + ρ_{ice}d g = 136 bar

d = (135 *10^{5}N/m^{2})/(917 kg/m^{3}* 9.8 m/s^{2}) = 1500 m.