If 90 g of molten lead at 327.3 ^{o}C is poured into a 300 g casting
form made of iron and initially at 20 ^{o}C, what is the final
temperature of the system? Assume no energy is lost to the environment.

Solution:

- Concepts:

Specific heat capacity:

c = ΔQ/(mΔT)

The specific heat capacity c is the amount of energy it takes to raise the temperature of one kg of material by 1 degree Kelvin or Celsius.

Latent heat of fusion: ΔQ = Lm

The latent heat is the heat released or absorbed per unit mass by a system in a reversible isobaric-isothermal change of phase. - Reasoning:

Energy is conserved. The energy released by the lead is equal to the energy absorbed by the iron. - Details of the calculation:

The melting point of lead is 327.3^{o }C.

Assume the final temperature of the system is T. Then the amount of energy released by the lead as it solidifies is

ΔQ = m_{lead}L_{lead}= 0.09 kg*(2.45*10^{4 }J/kg) = 2205 J,

and the amount of energy released as it cools is

ΔQ = m_{lead}c_{lead}ΔT = 0.09 kg*(128 J/(kg^{o}C))(327.3^{o}C - T) = (11.52 J/^{o}C)(327.3^{o}C - T).

This energy is absorbed by the iron. For the iron we therefore have

2205 J + (11.52 J/^{o}C) (327.3^{o}C - T) = m_{iron}c_{iron}ΔT = 0.3 kg*(448 J/(kg^{o}C))(T - 20^{o}C).

5975.5 J - (11.52 J/^{o}C)T = (134 J/^{o}C)T - 2688 J.

8663.5 J = (145.52 J/^{o}C)T.

T = 59.5^{o }C.

Pure water can be super-cooled at standard atmospheric pressure to below its
normal freezing point of 0 °C. Assume that a mass of water has been cooled as a
liquid to –5 °C, and a small (negligible mass) crystal of ice is introduced to
act as a “seed” or starting point of crystallization. If the subsequent change
of state occurs adiabatically and at constant pressure, what fraction of the
system solidifies? Assume the latent heat of fusion of the water is 80 kcal/kg
and that the specific heat of water is 1 kcal/(kg ^{o}C).

Solution:

- Concepts:

Specific heat, latent heat - Reasoning:

The latent heat released as some of the water freezes heats the system. - Details of the calculation:

The freezing stops when the ice and the water are at 0^{o}C. Let M be the mass of the water, M' the amount that turns to ice.

Energy conservation: M(1 kcal/(kg^{o}C))*(5^{o}C) = M'(80 kcal/kg).

M'/M = 5/80 = 1/16. 6.25% of the system solidifies.

A liquid of unknown specific heat at a temperature of 20°C was mixed with water at 80°C in a well-insulated container. The final temperature was measured to be 50°C, and the combined mass of the two liquids was measured to be 240-g. In a second experiment with both liquids at the same initial temperatures, 20-g less of the liquid of unknown specific heat was poured into the same amount of water as before. This time the equilibrium temperature was found to be 52°C. Determine the specific heat of the liquid. The specific heat of water is 4187 J/Kg°C or 1 kcal/kg°C.

Solution:

- Concepts:

Specific heat capacity:

c = ΔQ/(mΔT)

The specific heat capacity c is the amount of energy it takes to raise the temperature of one kg of material by 1 degree Kelvin or Celsius. - Reasoning:

Energy is conserved. Knowing the specific heat capacity of the liquids we can calculate how it is shared between the liquids. - Details of the calculation:

ΔQ = cmΔT

c_{l}m_{l}30 = c_{w}m_{w}30, c_{l}m_{l}= m_{w}, m_{l }+ m_{w}= 0.24

c_{l}(m_{l}– 0.02)32 = c_{w}m_{w}28

3 equations, 3 unknowns

m_{l }= 0.16

m_{w}= 0.08

c_{l}= 0.5 kcal/(kg^{o}C)

An electric coffee pot contains 2 liters of water which it heats from 20^{o}
C to boiling in 5 minutes. The supply voltage is 120V and each kWh costs 10
cents. Calculate

(a) the electric power converted,

(b) the cost of making ten pots of coffee,

(c) the resistance of the heating element, and

(d) the current in the element.

Solution:

- Concepts:

Specific heat, energy and power - Reasoning:

This problem asks us to recall simple physics concepts:. - Details of the calculation:

(a) Energy needed to heat the water: 1 kcal/(kg^{o}C) *80^{o}C * 2 kg = 160 kcal = 669760 J

Power = 669760 J/ (5*60 s) = 2232.5 W.

Each second 2232.5 J of electric energy are converted into thermal energy.

(b) 1 kWh = 1000 J/s*3600 s = 3.6*10^{6}J.

10*10 cents * 669760 J / 3.6*10^{6}J = 18.6 cents.

(c) P = IV = V^{2}/R. R = (120 V)^{2}/ 2232.5 Ω) = 6.45 Ω.

(d) I = 2232.5 W/(120 V) = 18.6 A.

How much thermal energy is required to change a 40 g ice cube from a solid at
-10 ^{o}C to steam at 110 ^{o}C?

Solution:

- Concepts:

Specific heat capacity:

c = ΔQ/(mΔT)

The specific heat capacity c is the amount of energy it takes to raise the temperature of one kg of material by 1 degree Kelvin or Celsius.

Latent heat: ΔQ = Lm

The latent heat is the heat released or absorbed per unit mass by a system in a reversible isobaric-isothermal change of phase. - Reasoning:

We are asked to find the amount of heat absorbed by a substance when its temperature increases and it changes phase twice. - Details of the calculation:

To raise the temperature of the ice to 0^{o}C we need

ΔQ = 0.04 kg*(0.49 kcal/(kg^{o}C))10^{ o}C = 0.196 kcal.

To melt the ice we need

ΔQ = 0.04 kg*80 kcal/kg = 3.2 kcal.

To raise the temperature of the water to 100^{o}C we need

ΔQ = 0.04 kg*(1 kcal/(kg^{o}C))100^{o}C= 4 kcal.

To boil the water we need

ΔQ = 0.04 kg*540 kcal/kg = 21.6 kcal.

To raise the temperature of the steam to 110^{o}C we need

ΔQ = 0.04kg*(0.48 kcal/(kg^{o}C))10^{o}C = 0.192 kcal.

The total thermal energy required is

(0.196 + 3.2 + 4 + 21.6 + 0.192)kcal = 29.188 kcal.

A capacitor, C = 100 μF, is charged to a potential of 25 kV. The
capacitor is then discharged through a 1 kW resistor immersed in and at
equilibrium with 500 ml of water. The water is at an initial temperature
of 20 ^{o}C. Find the final, equilibrium temperature of the water
(specific heat 4187 J/kg^{o}C), if the resistor has specific heat of 710
J/kg^{o}C and a mass of 100 g.

Solution:

- Concepts:

Energy conservation, specific heat - Reasoning:

The electrostatic potential energy stored in the capacitor will be converted into thermal energy - Details of the calculation:

Energy stored in the capacitor: ½CV^{2 }Increase in thermal energy: m_{water}c_{water}ΔT + m_{res}c_{res}ΔT

½CV^{2}= m_{water}c_{water}ΔT + m_{res}c_{res}ΔT

ΔT = ½CV^{2}/( m_{water}c_{water}+ m_{res}c_{res})

ΔT =14.4^{ o}C

Final equilibrium temperature: 34.4^{o}C.

In his short story "A Slight Case of Sunstroke", Arthur C. Clarke writes of a
stadium full of disgruntled soccer fans barbecuing the dishonest referee by
reflecting sunlight on him with mirrors found under their seats.

(a)
Imagine a stadium at the equator at noon (i.e. the sun's directly overhead),
with 50,000 fans. Assuming that sunlight delivers about 1000 watts per square
meter to the surface of the Earth, and that each fan is holding a 0.25 m^{2} mirror
at 45^{o}, how much power would be available to be projected onto a
dishonest referee?

(b) To be humane, let us replace the referee with a
50 kg cylinder of 37 ^{o}C water. Assuming this cylinder absorbs
all of the reflected light from the mirror - wielding fans, how long will it
take for it to reach 100 ^{o}C? (The heat capacity of water is
about 4200 J/(kg^{o}C).)

Solution:

- Concepts:

Specific heat capacity, power - Reasoning:

The total reflected power is the energy per unit time supplied to the water. The specific heat capacity lets us calculate how much energy is required to increase the temperature of the water by 63^{ o}C. - Details of the calculation:

(a) I_{sun}= 1000 W/m^{2}

N_{mirrors}= 50000

Since the sun is directly overhead, and the mirrors are tilted at 45^{o}, the area on each mirror exposed to the sun is really A_{mirror }= 0.25 m^{2}*cos45^{o}= 0.18 m^{2}.

Total reflected power:

P_{total}= I_{sun}*A_{mirror}*N_{mirror}= 8.839*10^{6 }W.

(b) 9*10^{6 }W of power are being supplied to a 50 kg cylinder of water. To find out the time it takes to get it to 100^{o}C, we first determine how much energy is required:

Q = mcΔT = 50 kg*4200 J/(kg^{o}C)*(100 - 37)^{ o}C = 1.3*10^{7 }J.

P = Q/t, t = Q/P = 1.3*10^{7 }J/(9*10^{6 }W) = 1.47 s.

In a solar collector, water flows through pipes that collect heat from an
area of 10 m^{2}. The collector faces the Sun and the intensity of
sunlight incident on it is 2000 W/m^{2}. At what rate (in kg/minute)
should the water circulate through the pipes so that it is heated by 40^{o}C,
if the collector efficiency is 30%? (The specific heat of water is c = 4200
J/(kg ^{o}C)).

Solution:

- Concepts:

Energy conservation, specific heat - 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 = (20000 J/s)*0.3/(4200 J/(kg^{o}C))/(40^{o}C) = 0.036 kg/s = 2.14 kg/minute

The water should circulate at 2.14 kg/min = 2.14 liter/min.

Phonons are quantized lattice vibrations, and many aspects of these
excitations can be understood in terms of simple mode counting.

(a)
Estimate the number of phonon modes in 1 cm^{3} of a crystalline
material with an inter-atomic spacing of 2 Angstrom.

(b) Assuming that
in thermal equilibrium each phonon mode has k_{B}T of energy, give a
numerical estimate of the heat capacity ΔE/ΔT of this 1 cm^{3} of
material, in [J/K].

Let C_{P} and C_{V} denote the molar specific heat of an
ideal gas at constant pressure and at constant volume, respectively. Show that
C_{P} - C_{V} = R.

Solution:

- Concepts:

Specific heat, internal energy, energy conservation, the ideal gas law - Reasoning:

We note that the internal energy of an ideal gas is proportional to its temperature. - Details of the calculation:

Definitions:

at constant pressure: ΔQ = nC_{P}ΔT

at constant volume: ΔQ = nC_{V}ΔT

Energy conservation:

at constant pressure: ΔU = ΔQ - ΔW = nC_{P}ΔT - PΔV = nC_{P}ΔT - nRΔT

at constant volume: ΔU = nC_{V}ΔT

ΔU depends only on the change in temperature, so nC_{P}ΔT - nRΔT = nC_{V}ΔT,

C_{P}- C_{V}= R.

Let C_{P} and C_{V} denote the molar specific heat of an
ideal diatomic gas at room temperature at constant pressure and constant volume,
respectively. Find the ratio C_{P}/C_{V}.

Solution:

- Concepts:

Specific heat, internal energy - Reasoning:

We note that the internal energy of an ideal gas is proportional to its temperature. - Details of the calculation:

at constant pressure: ΔQ = nC_{P}ΔT

at constant volume: ΔQ = nC_{V}ΔT

Energy conservation:

at constant pressure: ΔU = ΔQ - ΔW = nC_{P}ΔT - PΔV = nC_{P}ΔT - nRΔT

at constant volume: ΔU = nC_{V}ΔT

ΔU depends only on the change in temperature.

C_{P}/C_{V}= (ΔU + nRΔT)/ΔU =((5/2)nRΔT + nRΔT)/((5/2)nRΔT) = 7/5.

[We have U = n(5/2)RT for a diatomic gas at room temperature.

kT ~ 2.6*10^{-2}eV << hω ~ 2*10^{-1}eV, the vibrational mode cannot be excited at room temperature.]