Online Test — Thermal Properties of Matter
18 Questions • 15 min • Chapter MCQ
15:00
Question 1 of 18
Absolute zero corresponds to a Celsius temperature of:
$0\,^\circ$C
$100\,^\circ$C
$-273.15\,^\circ$C
$-100\,^\circ$C
Explanation: Absolute zero ($0$ K) is $-273.15\,^\circ$C, the lowest possible temperature.
Question 2 of 18
The Celsius and Fahrenheit scales read the same numerical value at:
$0\,^\circ$
$-40\,^\circ$
$40\,^\circ$
$100\,^\circ$
Explanation: Solving $C=\frac{5}{9}(C-32)$ gives $C=-40$, so $-40\,^\circ$C $=-40\,^\circ$F.
Question 3 of 18
The coefficient of volume expansion of an isotropic solid in terms of $\alpha$ is:
$\alpha$
$2\alpha$
$3\alpha$
$\frac{\alpha}{3}$
Explanation: $\gamma=3\alpha$ for an isotropic solid.
Question 4 of 18
Water has its maximum density at:
$0\,^\circ$C
$4\,^\circ$C
$100\,^\circ$C
$-4\,^\circ$C
Explanation: Due to anomalous expansion, water is densest at $4\,^\circ$C.
Question 5 of 18
The increase in length of a rod on heating is $\Delta L=$:
$\alpha\,\Delta T$
$\alpha L\,\Delta T$
$\frac{L\,\Delta T}{\alpha}$
$\alpha L$
Explanation: Linear expansion: $\Delta L=\alpha L\,\Delta T$.
Question 6 of 18
The amount of heat to raise the temperature of mass $m$ by $\Delta T$ is:
$Q=mL$
$Q=mc\,\Delta T$
$Q=\frac{m\,\Delta T}{c}$
$Q=c\,\Delta T$
Explanation: Without a change of state, $Q=mc\,\Delta T$.
Question 7 of 18
The specific heat of water is about:
$2100\ \text{J}\,\text{kg}^{-1}\,\text{K}^{-1}$
$4186\ \text{J}\,\text{kg}^{-1}\,\text{K}^{-1}$
$900\ \text{J}\,\text{kg}^{-1}\,\text{K}^{-1}$
$334\ \text{J}\,\text{kg}^{-1}\,\text{K}^{-1}$
Explanation: Water's specific heat is about $4186\ \text{J}\,\text{kg}^{-1}\,\text{K}^{-1}$, which is unusually high.
Question 8 of 18
During the change of state of a substance, its temperature:
increases
decreases
remains constant
fluctuates
Explanation: Heat goes into latent heat (bonds) during a change of state, so the temperature stays constant.
Question 9 of 18
The latent heat of fusion of ice is about:
$3.34\times10^{5}\ \text{J}\,\text{kg}^{-1}$
$22.6\times10^{5}\ \text{J}\,\text{kg}^{-1}$
$4186\ \text{J}\,\text{kg}^{-1}$
$5.67\times10^{-8}\ \text{J}\,\text{kg}^{-1}$
Explanation: $L_f$ of ice is about $3.34\times10^{5}\ \text{J}\,\text{kg}^{-1}$.
Question 10 of 18
The principle of calorimetry expresses the conservation of:
momentum
energy
charge
angular momentum
Explanation: Heat lost by the hot body equals heat gained by the cold body — conservation of energy.
Question 11 of 18
The heat required for a change of state is $Q=$:
$mc\,\Delta T$
$mL$
$\frac{m}{L}$
$L\,\Delta T$
Explanation: $Q=mL$, where $L$ is the latent heat.
Question 12 of 18
Heat transfer through a solid metal rod is mainly by:
convection
radiation
conduction
evaporation
Explanation: In solids, especially metals, heat travels by conduction (molecule to molecule and via free electrons).
Question 13 of 18
The rate of heat conduction through a rod is:
$\frac{kL\,\Delta T}{A}$
$\frac{kA\,\Delta T}{L}$
$\frac{kA L}{\Delta T}$
$kAL\,\Delta T$
Explanation: $\frac{Q}{t}=\frac{kA\,\Delta T}{L}$.
Question 14 of 18
Convection currents are set up because heated fluid becomes:
denser and sinks
less dense and rises
solid
stationary
Explanation: Heated fluid expands, becomes less dense and rises, while cooler fluid sinks to replace it.
Question 15 of 18
The mode of heat transfer that requires no medium is:
conduction
convection
radiation
all of these
Explanation: Radiation is carried by electromagnetic waves and needs no medium.
Question 16 of 18
According to the Stefan–Boltzmann law, $E$ is proportional to:
$T$
$T^2$
$T^3$
$T^4$
Explanation: $E=\sigma T^4$, the fourth power of the absolute temperature.
Question 17 of 18
Wien's displacement law states that $\lambda_m$ is:
directly proportional to $T$
inversely proportional to $T$
independent of $T$
proportional to $T^4$
Explanation: $\lambda_m T=b$, so the peak wavelength is inversely proportional to absolute temperature.
Question 18 of 18
Newton's law of cooling holds best when the temperature difference between body and surroundings is:
very large
small
zero
negative
Explanation: The law $\frac{dT}{dt}\propto(T-T_s)$ is valid for small temperature differences.