Specific Heat and Molar Heat Capacity: notes and previous year questions

Answer: 7

The amount of gas is fixed, so $\dfrac{p_{1}V_{1}}{T_{1}} = \dfrac{p_{2}V_{2}}{T_{2}}$, with the temperatures in kelvin: $27\ {}^{\circ}\mathrm{C} = 300\ \mathrm{K}$ and $77\ {}^{\circ}\mathrm{C} = 350\ \mathrm{K}$. The volumes can stay in $\mathrm{cm^{3}}$ and the pressures in atmospheres, since only ratios appear.

$p_{2} = p_{1} \times \dfrac{V_{1}}{V_{2}} \times \dfrac{T_{2}}{T_{1}} = 2 \times \dfrac{60}{20} \times \dfrac{350}{300} = 7\ \mathrm{atm}$

The trap is forgetting the heating: squeezing to a third of the volume alone gives $6\ \mathrm{atm}$, and the rise to $350\ \mathrm{K}$ adds the rest. Putting the Celsius values into the ratio, $\dfrac{77}{27}$, gives about $17\ \mathrm{atm}$ — far too much.

Answer: Option A

The two gases share one temperature, so in $v_{\mathrm{rms}} = \sqrt{\dfrac{3RT}{M}}$ only the molar mass differs, and $v_{\mathrm{rms}} \propto \dfrac{1}{\sqrt{M}}$:

$\dfrac{v_{\mathrm{rms}}^{\mathrm{Ar}}}{v_{\mathrm{rms}}^{\mathrm{Cl}}} = \sqrt{\dfrac{M_{\mathrm{Cl}}}{M_{\mathrm{Ar}}}} = \sqrt{\dfrac{70}{40}} = \sqrt{\dfrac{7}{4}} = \dfrac{\sqrt{7}}{2}$

The $2 : 1$ ratio of masses in the flask plays no part: how much of each gas there is does not change how fast its molecules move.

The trap is D, $\dfrac{2}{\sqrt{7}}$, the ratio upside down — it would make the heavier chlorine molecules the faster ones. B, $\dfrac{7}{4}$, forgets the square root, and C, $\dfrac{7}{2}$, lets the $2 : 1$ mass ratio in as well.

Answer: 25000

Here the specific heat is written in terms of the absolute temperature, so $T$ must be in kelvin: $-73\ {}^{\circ}\mathrm{C}$ is $200\ \mathrm{K}$ and $27\ {}^{\circ}\mathrm{C}$ is $300\ \mathrm{K}$. For $1\ \mathrm{kg}$,

$Q = \displaystyle\int_{200}^{300} kT\,dT = k\left[\dfrac{T^{2}}{2}\right]_{200}^{300} = k \times \dfrac{90\,000 - 40\,000}{2} = 25\,000\,k$

so $n = 25\,000$.

The trap is putting the Celsius values into $\dfrac{T^{2}}{2}$: $\dfrac{27^{2} - 73^{2}}{2}$ is negative, which no heat for warming can be. Another is using $c$ at a single temperature times the rise, $k \times 300 \times 100 = 30\,000\,k$, which forgets that $c$ is smaller at the start.

Answer: 30 g

$w = \dfrac{140 \times 900}{4200} = 30\ \mathrm{g}$. The can takes as much heat per kelvin as $30\ \mathrm{g}$ of water would: aluminium's specific heat is less than a quarter of water's, so its $140\ \mathrm{g}$ counts for much less water.

The trap is turning the fraction over, $\dfrac{140 \times 4200}{900} \approx 653\ \mathrm{g}$, which would make the can count for more water than its own mass — impossible for a material with a smaller specific heat than water. Another is answering $140\ \mathrm{g}$, as if the can were made of water.

Answer: Options A, C

A: the raised pressure lets the water get hotter than $100\ {}^{\circ}\mathrm{C}$ before it boils, and hotter water cooks faster. C is regelation: the ice melts under the wire and refreezes behind it, so the block closes up again.

B has the effect the wrong way round: the thinner air means lower pressure, which lowers the boiling point, so mountain water boils below $100\ {}^{\circ}\mathrm{C}$ — and cooks food more slowly. D is the trap: all three states coexist only at the triple point, one temperature and one pressure, not along a whole range.

Answer: 565.7 m/s

In kelvin the gas goes from $300\ \mathrm{K}$ to $600\ \mathrm{K}$, which doubles $T$. So $v_{\mathrm{rms}} = 400 \times \sqrt{2} \approx 566\ \mathrm{m/s}$.

The trap is the Celsius ratio: $400 \times \sqrt{\dfrac{327}{27}} \approx 1392\ \mathrm{m/s}$. Another is forgetting the square root and doubling the speed to $800\ \mathrm{m/s}$.