AP Physics

Unit 1 - Vectors and Kinematics

MCQ
Mathematical
Intermediate

Pro

Pro

Educator

Upgrade For More Credits
0
Step Reasoning
Identify the target equation for latent heat of fusion from the phase-change portion of the graph.
\[L_f = \dfrac{\Delta Q_{\text{melt}}}{m}\]
The question asks for the latent heat of fusion, which occurs at constant temperature during melting (the horizontal region of the graph).
Calculate the latent heat of fusion using the energy interval of the horizontal line segment.
\[\Delta Q_{\text{melt}} = 140\text{ kJ} – 60\text{ kJ} = 80\text{ kJ}\]
\[L_f = \dfrac{80\text{ kJ}}{0.50\text{ kg}} = 160\text{ kJ/kg}\]
Melting begins at \(Q = 60\text{ kJ}\) and ends at \(Q = 140\text{ kJ}\), so the energy added during melting is the difference between these values.
Relate the slope of the \(T\)-versus-\(Q\) graph to specific heat capacity.
\[c = \dfrac{\Delta Q}{m \Delta T} = \dfrac{1}{m \left(\dfrac{\Delta T}{\Delta Q}\right)} = \dfrac{1}{m \cdot \text{slope}}\]
To compare \(c_{\text{solid}}\) and \(c_{\text{liquid}}\), express \(c\) in terms of the graph’s variables using \(Q = m c \Delta T\).
Compare the slopes of the solid and liquid heating regions to determine the relative specific heat capacities.
\[\text{slope}_{\text{solid}} = \dfrac{80\text{ }^\circ\text{C} – 20\text{ }^\circ\text{C}}{60\text{ kJ} – 0\text{ kJ}} = 1.0\text{ }^\circ\text{C/kJ}\]
\[\text{slope}_{\text{liquid}} = \dfrac{140\text{ }^\circ\text{C} – 80\text{ }^\circ\text{C}}{260\text{ kJ} – 140\text{ kJ}} = \dfrac{60}{120} = 0.5\text{ }^\circ\text{C/kJ}\]
\[\text{Since } \text{slope}_{\text{liquid}} < \text{slope}_{\text{solid}}, \text{ we have } c_{\text{solid}} < c_{\text{liquid}}.\]
Since \(c\) is inversely proportional to the slope, a flatter line (smaller slope) corresponds to a larger specific heat capacity.

Why each choice is correct or incorrect:

(A) Uses \(\Delta Q_{\text{melt}} = 80\text{ kJ}\) directly without dividing by mass \(m = 0.50\text{ kg}\), and confuses a steeper slope with a larger specific heat capacity.

(B) Uses the total cumulative heat at the end of melting (\(140\text{ kJ}\)) instead of the thermal energy added during the phase change (\(80\text{ kJ}\)) to find \(L_f = 140 / 0.50 = 280\text{ kJ/kg}\).

(C) This is the correct answer. Correctly calculates \(L_f = 80\text{ kJ} / 0.50\text{ kg} = 160\text{ kJ/kg}\) and recognizes that a smaller slope in the liquid region means \(c_{\text{liquid}} > c_{\text{solid}}\).

(D) Correctly calculates \(L_f = 160\text{ kJ/kg}\), but incorrectly assumes specific heat capacity is directly proportional to slope rather than inversely proportional.

Need Help? Ask Phy To Explain

A Major Upgrade To Phy Is Coming Soon — Stay Tuned

Just Drag and Drop!
Quick Actions ?
×

NEW UBQ QUIZ LAB

100s of AP aligned questions and quizzes to help you get a 5 even faster. Full Mock exams with Auto Grading and Adaptive explanations. Try out Nerd Notes', state of the art, quiz platform.

Topics in this question

We'll help clarify entire units in one hour or less — guaranteed.

A self paced course with videos, problems sets, and everything you need to get a 5. Trusted by over 15k students and over 200 schools.

Go Pro to remove ads + unlimited access to our AI learning tools.

C

Nerd Notes

Discover the world's best Physics resources

Continue with

By continuing you (1) agree to our Terms of Use and Terms of Sale and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.

Error Report

Sign in before submitting feedback.

KinematicsForces
\(\Delta x = v_i t + \frac{1}{2} at^2\)\(F = ma\)
\(v = v_i + at\)\(F_g = \frac{G m_1 m_2}{r^2}\)
\(v^2 = v_i^2 + 2a \Delta x\)\(f = \mu N\)
\(\Delta x = \frac{v_i + v}{2} t\)\(F_s =-kx\)
\(v^2 = v_f^2 \,-\, 2a \Delta x\) 
Circular MotionEnergy
\(F_c = \frac{mv^2}{r}\)\(KE = \frac{1}{2} mv^2\)
\(a_c = \frac{v^2}{r}\)\(PE = mgh\)
\(T = 2\pi \sqrt{\frac{r}{g}}\)\(KE_i + PE_i = KE_f + PE_f\)
 \(W = Fd \cos\theta\)
MomentumTorque and Rotations
\(p = mv\)\(\tau = r \cdot F \cdot \sin(\theta)\)
\(J = \Delta p\)\(I = \sum mr^2\)
\(p_i = p_f\)\(L = I \cdot \omega\)
Simple Harmonic MotionFluids
\(F = -kx\)\(P = \frac{F}{A}\)
\(T = 2\pi \sqrt{\frac{l}{g}}\)\(P_{\text{total}} = P_{\text{atm}} + \rho gh\)
\(T = 2\pi \sqrt{\frac{m}{k}}\)\(Q = Av\)
\(x(t) = A \cos(\omega t + \phi)\)\(F_b = \rho V g\)
\(a = -\omega^2 x\)\(A_1v_1 = A_2v_2\)
ConstantDescription
[katex]g[/katex]Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface
[katex]G[/katex]Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex]
[katex]\mu_k[/katex] and [katex]\mu_s[/katex]Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion.
[katex]k[/katex]Spring constant, in [katex]\text{N/m}[/katex]
[katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex]Mass of the Earth
[katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex]Mass of the Moon
[katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex]Mass of the Sun
VariableSI Unit
[katex]s[/katex] (Displacement)[katex]\text{meters (m)}[/katex]
[katex]v[/katex] (Velocity)[katex]\text{meters per second (m/s)}[/katex]
[katex]a[/katex] (Acceleration)[katex]\text{meters per second squared (m/s}^2\text{)}[/katex]
[katex]t[/katex] (Time)[katex]\text{seconds (s)}[/katex]
[katex]m[/katex] (Mass)[katex]\text{kilograms (kg)}[/katex]
VariableDerived SI Unit
[katex]F[/katex] (Force)[katex]\text{newtons (N)}[/katex]
[katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy)[katex]\text{joules (J)}[/katex]
[katex]P[/katex] (Power)[katex]\text{watts (W)}[/katex]
[katex]p[/katex] (Momentum)[katex]\text{kilogram meters per second (kgm/s)}[/katex]
[katex]\omega[/katex] (Angular Velocity)[katex]\text{radians per second (rad/s)}[/katex]
[katex]\tau[/katex] (Torque)[katex]\text{newton meters (Nm)}[/katex]
[katex]I[/katex] (Moment of Inertia)[katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex]
[katex]f[/katex] (Frequency)[katex]\text{hertz (Hz)}[/katex]

Metric Prefixes

Example of using unit analysis: Convert 5 kilometers to millimeters. 

  1. Start with the given measurement: [katex]\text{5 km}[/katex]

  2. Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]

  3. Perform the multiplication: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}[/katex]

  4. Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

[katex]10^{-12}[/katex]

Nano-

n

[katex]10^{-9}[/katex]

Micro-

µ

[katex]10^{-6}[/katex]

Milli-

m

[katex]10^{-3}[/katex]

Centi-

c

[katex]10^{-2}[/katex]

Deci-

d

[katex]10^{-1}[/katex]

(Base unit)

[katex]10^{0}[/katex]

Deca- or Deka-

da

[katex]10^{1}[/katex]

Hecto-

h

[katex]10^{2}[/katex]

Kilo-

k

[katex]10^{3}[/katex]

Mega-

M

[katex]10^{6}[/katex]

Giga-

G

[katex]10^{9}[/katex]

Tera-

T

[katex]10^{12}[/katex]

Sign In to View Your Questions

Share This Question

Enjoying UBQ? Share the 🔗 with friends!

Link Copied!

PRO TIER

One price to unlock most advanced version of Phy across all our tools.

$20

per month

Billed Monthly. Cancel Anytime.

We use cookies to improve your experience. By continuing to browse on Nerd Notes, you accept the use of cookies as outlined in our privacy policy.

Physics Shouldn't Be Hard

So I designed the Ultimate AP Physics 1 Course so you can learn faster and score higher in less than 100 hours.
Trusted by 10k+ Students

📚 Predict Your AP Physics Exam Score

Try our free calculator to see what you need to get a 5 on the 2026 AP Physics 1 exam.

Feeling uneasy about your next physics test? We'll boost your grade in 3 lessons or less—guaranteed