Week 10: Thermal Physics

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[1] Overview⚓︎

I recommend reviewing the following concepts, they will probably come up in recitation:

Temperature-scale conversions (Kelvin $\leftrightarrow$ Celsius $\leftrightarrow$ Fahrenheit).
Linear thermal expansion ($\Delta L = \alpha L_0 \Delta T$).
Calorimetry with phase changes (melting/freezing/boiling/condensing, and keeping track of heat gained vs. heat lost).
Heating curves / multi-step energy accounting (warming a substance, changing phase, then warming again).
Power and heating time (using $P = Q/t$ to connect energy input and elapsed time).
Energy conversion into thermal energy (for example, kinetic energy turning into internal energy and causing melting).

[2] Practice Problems⚓︎

If you're comfortable with the "★★★" problems, you should get 100 on the recitation.

If you want to print these questions, simply press Ctrl + P while on this page, and it should come out formatted nicely.

Difficulty key:

★☆☆ = beginner
★★☆ = standard
★★★ = challenging / multi-step

Water/Ice/Steam Constants

$c_{\mathrm{water}} = 4186 \ \mathrm{J/(kg \cdot ^\circ C)}$
$c_{\mathrm{ice}} = 2090 \ \mathrm{J/(kg \cdot ^\circ C)}$
$L_f = 3.34 \times 10^5 \ \mathrm{J/kg}$
$L_v = 2.26 \times 10^6 \ \mathrm{J/kg}$

[2.1] Temperature Scales and Thermal Expansion⚓︎

★☆☆
A vaccine cooler warms from $41^\circ \mathrm{F}$ to $86^\circ \mathrm{F}$ during transport. Find the temperature change in Celsius, as well as Kelvin.

Solution

Temperature differences convert by $\Delta T_C = \frac{5}{9}\Delta T_F$.

Here $\Delta T_F = 86 - 41 = 45^\circ \mathrm{F}$, so

\[\Delta T_C = \frac{5}{9}(45)=25.0^\circ \mathrm{C}.\]

A temperature change in Kelvin has the same numerical value as in Celsius, so

\[\Delta T_K = 25.0 \ \mathrm{K}.\]

Answer: $25.0^\circ \mathrm{C}$ and $25.0 \ \mathrm{K}$.
★★☆
A $2.40 \ \mathrm{m}$ aluminum bar is at $18^\circ \mathrm{C}$ in the morning. By afternoon it reaches $63^\circ \mathrm{C}$. How much does its length increase?
- $\alpha _{\mathrm{Al}} = 24 \times 10^{-6} \ \left( ^\circ \mathrm{C} \right)^{-1}$.
Solution

Use linear expansion:

\[\Delta L = \alpha L_0 \Delta T.\]

Here $\Delta T = 63 - 18 = 45^\circ \mathrm{C}$, so

\[\Delta L = \left( 24 \times 10^{-6} \right)(2.40)(45)=2.59 \times 10^{-3} \ \mathrm{m}.\]

This is $2.59 \ \mathrm{mm}$.

Answer: $\Delta L = 2.59 \times 10^{-3} \ \mathrm{m} = 2.59 \ \mathrm{mm}$.
★★★
A brass ring has an inner diameter of $8.000 \ \mathrm{cm}$ at $20.0^\circ \mathrm{C}$. To what temperature must it be heated so that its inner diameter becomes $8.012 \ \mathrm{cm}$?
- $\alpha _{Brass} = 19 \times 10^{-6} \ \left( ^\circ \mathrm{C} \right)^{-1}$.
Solution

The hole expands just like the brass, so

\[\Delta D = \alpha D_0 \Delta T.\]

Here $\Delta D = 8.012 - 8.000 = 0.012 \ \mathrm{cm}$, so

\[\Delta T = \frac{\Delta D}{\alpha D_0}=\frac{0.012}{\left( 19 \times 10^{-6} \right)(8.000)}=78.9^\circ \mathrm{C}.\]

Therefore,

\[T_f = 20.0^\circ \mathrm{C} + 78.9^\circ \mathrm{C} = 98.9^\circ \mathrm{C}.\]

Answer: $98.9^\circ \mathrm{C}$.

[2.2] Calorimetry Without and With Material Changes⚓︎

★☆☆
A $0.200 \ \mathrm{kg}$ copper block at $180^\circ \mathrm{C}$ is dropped into $0.500 \ \mathrm{kg}$ of water at $20.0^\circ \mathrm{C}$ in an insulated container. What is the final equilibrium temperature?
- $c_{\mathrm{Cu}} = 385 \ \mathrm{J/(kg \cdot ^\circ C)}$.
Solution

Since the container is insulated, heat lost by copper equals heat gained by water:

\[m_{\mathrm{Cu}} c_{\mathrm{Cu}} (180 - T_f) = m_w c_{\mathrm{water}} (T_f - 20.0).\]

Using $c_{\mathrm{water}} = 4186 \ \mathrm{J/(kg \cdot ^\circ C)}$,

\[0.200(385)(180 - T_f)=0.500(4186)(T_f - 20.0).\]

Solving gives

\[T_f = 25.7^\circ \mathrm{C}.\]

Answer: $25.7^\circ \mathrm{C}$.
★★☆
An aluminum calorimeter cup of mass $0.150 \ \mathrm{kg}$ contains $0.250 \ \mathrm{kg}$ of water at $22.0^\circ \mathrm{C}$. A $0.080 \ \mathrm{kg}$ brass sample initially at $95.0^\circ \mathrm{C}$ is placed into the cup. Find the final equilibrium temperature.
- $c_{\mathrm{Al}} = 900 \ \mathrm{J/(kg \cdot ^\circ C)}$
- $c_{\mathrm{brass}} = 380 \ \mathrm{J/(kg \cdot ^\circ C)}$.
Solution

The brass cools, while the water and aluminum cup warm up. Set heat lost equal to heat gained:

\[m_b c_b (95.0 - T_f) = m_w c_{\mathrm{water}} (T_f - 22.0) + m_{\mathrm{Al}} c_{\mathrm{Al}} (T_f - 22.0).\]

Using $c_{\mathrm{water}} = 4186 \ \mathrm{J/(kg \cdot ^\circ C)}$,

\[0.080(380)(95.0 - T_f)=0.250(4186)(T_f - 22.0)+0.150(900)(T_f - 22.0).\]

Solving gives

\[T_f = 23.8^\circ \mathrm{C}.\]

Answer: $23.8^\circ \mathrm{C}$.
★★★
A $3.0 \ \mathrm{kg}$ iron pan is at $250^\circ \mathrm{C}$. Into it, we pour $0.40 \ \mathrm{kg}$ of water initially at $20^\circ \mathrm{C}$. Assume atmospheric pressure and that the system is insulated from the surroundings. Determine the final state of the water, as well as the final temperature of the system.
- $c_{\mathrm{iron}} = 450 \ \mathrm{J/(kg \cdot ^\circ C)}$.
Solution

First check whether the iron can heat the water all the way to $100^\circ \mathrm{C}$.

Heat the water from $20^\circ \mathrm{C}$ to $100^\circ \mathrm{C}$:

\[Q_1 = m_w c_{\mathrm{water}} \Delta T = 0.40(4186)(80)=1.34 \times 10^5 \ \mathrm{J}.\]

Heat available if the iron cools from $250^\circ \mathrm{C}$ to $100^\circ \mathrm{C}$:

\[Q_{\mathrm{iron}} = m_{\mathrm{iron}} c_{\mathrm{iron}} \Delta T = 3.0(450)(150)=2.03 \times 10^5 \ \mathrm{J}.\]

Since $Q_{\mathrm{iron}} > Q_1$, the water reaches $100^\circ \mathrm{C}$ and some of it boils.

The leftover energy for vaporization is

\[Q_{\mathrm{left}} = 2.03 \times 10^5 - 1.34 \times 10^5 = 6.85 \times 10^4 \ \mathrm{J}.\]

So the mass that boils away is

\[m_{\mathrm{boiled}} = \frac{Q_{\mathrm{left}}}{L_v}=\frac{6.85 \times 10^4}{2.26 \times 10^6}=3.03 \times 10^{-2} \ \mathrm{kg}.\]

Answer: The final temperature is $100^\circ \mathrm{C}$, and the water ends as a mixture of liquid water and steam. About $0.030 \ \mathrm{kg}$ becomes steam, so about $0.370 \ \mathrm{kg}$ remains liquid.

[2.3] Phase Change and Energy Accounting⚓︎

★★☆
How much energy is required to convert $0.150 \ \mathrm{kg}$ of ice at $-12^\circ \mathrm{C}$ into liquid water at $25^\circ \mathrm{C}$?
Solution

This takes three steps:
1. Warm the ice from $-12^\circ \mathrm{C}$ to $0^\circ \mathrm{C}$.
2. Melt the ice.
3. Warm the liquid water from $0^\circ \mathrm{C}$ to $25^\circ \mathrm{C}$.
So

\[Q = m c_{\mathrm{ice}} (12) + m L_f + m c_{\mathrm{water}} (25).\]

Plugging in,

\[Q = (0.150)(2090)(12) + (0.150)(3.34 \times 10^5) + (0.150)(4186)(25).\]

\[Q = 6.96 \times 10^4 \ \mathrm{J}.\]

Answer: $6.96 \times 10^4 \ \mathrm{J}$.
★★☆
A picnic cooler contains $0.60 \ \mathrm{kg}$ of lemonade (assume the same specific heat as water) at $28^\circ \mathrm{C}$. How much ice at $0^\circ \mathrm{C}$ must be added so that the final temperature is $10^\circ \mathrm{C}$, with all the ice melted?

Solution

Heat lost by the lemonade equals heat gained by the ice.

The lemonade cools from $28^\circ \mathrm{C}$ to $10^\circ \mathrm{C}$:

\[Q_{\mathrm{lost}} = m c_{\mathrm{water}} \Delta T = (0.60)(4186)(28 - 10)=4.52 \times 10^4 \ \mathrm{J}.\]

Each kilogram of ice must melt and then warm from $0^\circ \mathrm{C}$ to $10^\circ \mathrm{C}$:

\[Q_{\mathrm{gain}} = m_i L_f + m_i c_{\mathrm{water}} (10).\]

Set them equal:

\[4.52 \times 10^4 = m_i \left[ 3.34 \times 10^5 + 4186(10) \right].\]

Solving gives

\[m_i = 0.120 \ \mathrm{kg}.\]

Answer: $0.120 \ \mathrm{kg}$ of ice.
★★★
In an insulated container, $0.20 \ \mathrm{kg}$ of ice at $-15^\circ \mathrm{C}$ is mixed with $0.50 \ \mathrm{kg}$ of liquid water at $60^\circ \mathrm{C}$. Find the final temperature, as well as whether any ice remains.

Solution

First see whether the warm water can completely melt the ice.

To warm the ice to $0^\circ \mathrm{C}$ and melt it:

\[Q_{\mathrm{need}} = m_{\mathrm{ice}} c_{\mathrm{ice}} (15) + m_{\mathrm{ice}} L_f.\]

\[Q_{\mathrm{need}} = (0.20)(2090)(15) + (0.20)(3.34 \times 10^5)=7.31 \times 10^4 \ \mathrm{J}.\]

Heat the warm water can release by cooling to $0^\circ \mathrm{C}$:

\[Q_{\mathrm{water}} = (0.50)(4186)(60)=1.26 \times 10^5 \ \mathrm{J}.\]

Since $Q_{\mathrm{water}} > Q_{\mathrm{need}}$, all the ice melts.

The leftover energy is

\[Q_{\mathrm{left}} = 1.26 \times 10^5 - 7.31 \times 10^4 = 5.25 \times 10^4 \ \mathrm{J}.\]

This warms the total $0.70 \ \mathrm{kg}$ of liquid water from $0^\circ \mathrm{C}$ to $T_f$:

\[Q_{\mathrm{left}} = (0.70)(4186)T_f.\]

So

\[T_f = 17.9^\circ \mathrm{C}.\]

Answer: The final temperature is $17.9^\circ \mathrm{C}$, and no ice remains.

[2.4] Steam, Power, and Mechanical Energy Turning Into Heat⚓︎

★★★
At atmospheric pressure, a mass $m_s$ of steam at $110^\circ \mathrm{C}$ is bubbled into a mass $m_w$ of liquid water at $15^\circ \mathrm{C}$ inside an insulated container. The final state is all liquid water at $50^\circ \mathrm{C}$. Find which substance loses heat and which gains heat, an expression for the heat lost by the steam, an expression for the heat gained by the liquid water, and the ratio $m_s/m_w$.
- $c_{\mathrm{steam}} = 2020 \ \mathrm{J/(kg \cdot ^\circ C)}$
- $c_{\mathrm{water}} = 4186 \ \mathrm{J/(kg \cdot ^\circ C)}$
- $L_v = 2.26 \times 10^6 \ \mathrm{J/kg}$
Solution

The steam loses heat, and the cooler liquid water gains heat.

The steam loses heat in three parts:
1. Cool steam from $110^\circ \mathrm{C}$ to $100^\circ \mathrm{C}$.
2. Condense the steam at $100^\circ \mathrm{C}$.
3. Cool the condensed water from $100^\circ \mathrm{C}$ to $50^\circ \mathrm{C}$.
So the heat lost by the steam is

\[Q_{\mathrm{steam}} = m_s c_{\mathrm{steam}} (110 - 100) + m_s L_v + m_s c_{\mathrm{water}} (100 - 50).\]

The liquid water warms from $15^\circ \mathrm{C}$ to $50^\circ \mathrm{C}$, so

\[Q_{\mathrm{water}} = m_w c_{\mathrm{water}} (50 - 15).\]

Energy conservation gives $Q_{\mathrm{steam}} = Q_{\mathrm{water}}$, so

\[\frac{m_s}{m_w}=\frac{c_{\mathrm{water}} (50 - 15)}{c_{\mathrm{steam}} (110 - 100) + L_v + c_{\mathrm{water}} (100 - 50)}.\]

Numerically,

\[\frac{m_s}{m_w}=\frac{4186(35)}{2020(10) + 2.26 \times 10^6 + 4186(50)}=0.0589.\]

Answer: Steam loses heat, liquid water gains heat, and $\displaystyle \frac{m_s}{m_w}=0.0589$.
★★☆
A $1.50 \ \mathrm{kW}$ immersion heater is used to turn $2.00 \ \mathrm{kg}$ of ice at $-20^\circ \mathrm{C}$ into steam at $100^\circ \mathrm{C}$. Assume no heat loss to the surroundings. How long does the entire process take?
Solution

The heater must:
1. Warm the ice from $-20^\circ \mathrm{C}$ to $0^\circ \mathrm{C}$.
2. Melt the ice.
3. Warm the water from $0^\circ \mathrm{C}$ to $100^\circ \mathrm{C}$.
4. Vaporize the water at $100^\circ \mathrm{C}$.
Total energy required:

\[Q = m c_{\mathrm{ice}} (20) + m L_f + m c_{\mathrm{water}} (100) + m L_v.\]

\[Q = (2.00)(2090)(20) + (2.00)(3.34 \times 10^5) + (2.00)(4186)(100) + (2.00)(2.26 \times 10^6).\]

\[Q = 6.11 \times 10^6 \ \mathrm{J}.\]

Since $P = Q/t$,

\[t=\frac{Q}{P}=\frac{6.11 \times 10^6}{1.50 \times 10^3}=4.07 \times 10^3 \ \mathrm{s}.\]

This is about $67.9 \ \mathrm{min}$.

Answer: $4.07 \times 10^3 \ \mathrm{s}$, or about $67.9 \ \mathrm{min}$.
★★★
A $78 \ \mathrm{kg}$ cyclist plus bike is moving on level ground and brakes to rest. A single $0.65 \ \mathrm{kg}$ steel brake rotor absorbs $60\%$ of the lost kinetic energy and warms from $22^\circ \mathrm{C}$ to $180^\circ \mathrm{C}$. Estimate the cyclist’s initial speed.
- $c_{\mathrm{steel}} = 460 \ \mathrm{J/(kg \cdot ^\circ C)}$.
Solution

The rotor gains thermal energy

\[Q = m c \Delta T = (0.65)(460)(180 - 22)=4.72 \times 10^4 \ \mathrm{J}.\]

This is $60\%$ of the lost kinetic energy, so

\[0.60 \left( \frac12 M v^2 \right)=4.72 \times 10^4.\]

With $M = 78 \ \mathrm{kg}$,

\[0.60 \left( \frac12 \right)(78)v^2 = 4.72 \times 10^4.\]

Solving,

\[v = 44.9 \ \mathrm{m/s}.\]

Answer: $44.9 \ \mathrm{m/s}$.