Temperature and Heat — Hard Practice Quiz
A Physics cheat sheet for Temperature and Heat — every key formula with its symbols defined — plus a hard-level practice quiz to test recall.
Formulas & key concepts
Fahrenheit to Celsius: Converts Celsius temperature \(T_C\) to Fahrenheit \(T_F\).
Kelvin to Celsius: Converts Celsius temperature \(T_C\) to Kelvin \(T\).
Linear Thermal Expansion: Change in length \(\Delta L\) due to temperature change \(\Delta T\).
Where: \(\alpha\) = coefficient of linear expansion
Volume Thermal Expansion: Change in volume \(\Delta V\) due to temperature change \(\Delta T\).
Where: \(\beta\) = coefficient of volume expansion
Heat Supplied/Removed (Specific Heat): Heat \(Q\) required to change temperature of mass \(m\) by \(\Delta T\).
Where: \(c\) = specific heat capacity
Heat Supplied/Removed (Latent Heat): Heat \(Q\) required for phase change of mass \(m\).
Where: \(L\) = latent heat
Practice quiz
A steel rod has an initial length of $2.000 \text{ m}$ at $68^\circ \text{F}$. If its temperature is raised to $300 \text{ K}$, what is its final length? The coefficient of linear expansion for steel is $\alpha = 1.2 \times 10^{-5} \text{ (C}^\circ)^{-1}$.
- $2.000164 \text{ m}$
- $2.000082 \text{ m}$
- $2.000246 \text{ m}$
- $2.000000 \text{ m}$
Answer: $2.000164 \text{ m}$
A $500 \text{ mL}$ sample of a liquid with a density of $0.80 \text{ g/cm}^3$ at $20^\circ \text{C}$ is heated to $70^\circ \text{C}$. If its specific heat capacity is $2.5 \text{ J/(g} \cdot \text{C}^\circ)$, and its coefficient of volume expansion is $\beta = 9.0 \times 10^{-4} \text{ (C}^\circ)^{-1}$, how much heat was supplied and what is the final volume of the liquid?
- $Q = 50 \text{ kJ}$, $V_f = 522.5 \text{ mL}$
- $Q = 50 \text{ kJ}$, $V_f = 500.0 \text{ mL}$
- $Q = 40 \text{ kJ}$, $V_f = 522.5 \text{ mL}$
- $Q = 40 \text{ kJ}$, $V_f = 500.0 \text{ mL}$
Answer: $Q = 50 \text{ kJ}$, $V_f = 522.5 \text{ mL}$
$100 \text{ g}$ of ice at $-10^\circ \text{C}$ is heated until it completely melts and its temperature rises to $20^\circ \text{C}$. Calculate the total heat required. Given: specific heat of ice $c_{\text{ice}} = 2100 \text{ J/(kg} \cdot \text{C}^\circ)$, specific heat of water $c_{\text{water}} = 4186 \text{ J/(kg} \cdot \text{C}^\circ)$, latent heat of fusion of ice $L_f = 3.34 \times 10^5 \text{ J/kg}$.
- $43.87 \text{ kJ}$
- $35.50 \text{ kJ}$
- $41.57 \text{ kJ}$
- $39.70 \text{ kJ}$
Answer: $43.87 \text{ kJ}$
Two rods, A and B, are made of different materials. Rod A has an initial length $L_{0A}$. Rod B has an initial length $L_{0B} = 2L_{0A}$. If both rods are heated by the same temperature change $\Delta T$, and their change in length is equal ($\Delta L_A = \Delta L_B$), what is the ratio $\alpha_A / \alpha_B$?
- $2$
- $1/2$
- $1$
- $4$
Answer: $2$
A liquid has an initial volume $V_0$ and density $\rho_0$ at temperature $T_0$. If its temperature increases by $\Delta T$, and its coefficient of volume expansion is $\beta$, what is its new density $\rho_f$ in terms of $\rho_0$, $\beta$, and $\Delta T$? Assume mass remains constant.
- $\rho_f = \rho_0 / (1 + \beta \Delta T)$
- $\rho_f = \rho_0 (1 + \beta \Delta T)$
- $\rho_f = \rho_0 (1 - \beta \Delta T)$
- $\rho_f = \rho_0 / (1 - \beta \Delta T)$
Answer: $\rho_f = \rho_0 / (1 + \beta \Delta T)$
$200 \text{ g}$ of a substance at its melting point ($0^\circ \text{C}$) is supplied with $60 \text{ kJ}$ of heat. If its latent heat of fusion is $L_f = 2.5 \times 10^5 \text{ J/kg}$ and its specific heat capacity in liquid phase is $c = 2000 \text{ J/(kg} \cdot \text{C}^\circ)$, what is the final temperature of the substance?
- $25^\circ \text{C}$
- $0^\circ \text{C}$
- $10^\circ \text{C}$
- $50^\circ \text{C}$
Answer: $25^\circ \text{C}$
A metal rod expands by $\Delta L$ when its temperature increases by $25^\circ \text{C}$. If the same rod were subjected to a temperature increase of $25^\circ \text{F}$, what would be its new change in length, $\Delta L'$?
- $\frac{5}{9} \Delta L$
- $\frac{9}{5} \Delta L$
- $\Delta L$
- $\frac{1}{25} \Delta L$
Answer: $\frac{5}{9} \Delta L$
A $1.0 \text{ kg}$ block of an unknown material has an initial volume of $5.0 \times 10^{-4} \text{ m}^3$ at $20^\circ \text{C}$. When $10 \text{ kJ}$ of heat is supplied to it, its temperature rises to $40^\circ \text{C}$ and its volume increases by $1.5 \times 10^{-6} \text{ m}^3$. Determine the specific heat capacity $c$ and the coefficient of volume expansion $\beta$ for this material.
- $c = 500 \text{ J/(kg} \cdot \text{C}^\circ)$, $\beta = 1.5 \times 10^{-4} \text{ (C}^\circ)^{-1}$
- $c = 250 \text{ J/(kg} \cdot \text{C}^\circ)$, $\beta = 3.0 \times 10^{-4} \text{ (C}^\circ)^{-1}$
- $c = 500 \text{ J/(kg} \cdot \text{C}^\circ)$, $\beta = 3.0 \times 10^{-4} \text{ (C}^\circ)^{-1}$
- $c = 250 \text{ J/(kg} \cdot \text{C}^\circ)$, $\beta = 1.5 \times 10^{-4} \text{ (C}^\circ)^{-1}$
Answer: $c = 500 \text{ J/(kg} \cdot \text{C}^\circ)$, $\beta = 1.5 \times 10^{-4} \text{ (C}^\circ)^{-1}$
Derive an expression for a temperature change in Kelvin ($\Delta T_K$) in terms of a temperature change in Fahrenheit ($\Delta T_F$).
- $\Delta T_K = \frac{5}{9} \Delta T_F$
- $\Delta T_K = \frac{9}{5} \Delta T_F$
- $\Delta T_K = \Delta T_F - 273.15$
- $\Delta T_K = \frac{5}{9} (\Delta T_F - 32)$
Answer: $\Delta T_K = \frac{5}{9} \Delta T_F$
A cube of an isotropic material has an initial side length $L_0$. If its temperature increases by $\Delta T$, what is the approximate relationship between its coefficient of linear expansion $\alpha$ and its coefficient of volume expansion $\beta$? Assume $\alpha \Delta T \ll 1$.
- $\beta = 3\alpha$
- $\beta = \alpha^3$
- $\beta = \alpha$
- $\beta = 2\alpha$
Answer: $\beta = 3\alpha$
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