Thermodynamics in Physics Questions and Answers

Answer: A

The temperature of the mixture is \(\dfrac{5T_{1} + 6T_{2}}{11}\).

In an insulated system the heat lost by the hotter liquid equals the heat gained by the cooler one. With equal masses \(m\) and final temperature \(T\):

\[ \begin{aligned} mS(T_{1} - T) &= m(1.2S)(T - T_{2}) \\ T_{1} + 1.2T_{2} &= 2.2T \\ T &= \frac{5T_{1} + 6T_{2}}{11} \end{aligned} \]

The liquid with the larger specific heat (the cooler one) has more weight in the average, so \(T_{2}\) gets coefficient 6. \(\frac{6T_{1} + 5T_{2}}{11}\) swaps the weights, and \(\frac{T_{1} + T_{2}}{2}\) would be right only for equal specific heats.

Answer: A

The heat supplied is \(3RT\).

At constant volume no work is done, and for a monatomic gas \(C_{V} = \frac{3}{2}R\) per mole:

\[ \begin{aligned} Q &= nC_{V}\Delta T = 1 \times \tfrac{3}{2}R \times (3T - T) \\ &= 3RT \end{aligned} \]

\(\frac{3}{2}RT\) forgets that the rise is \(2T\), not \(T\). \(5RT\) uses \(C_{P} = \frac{5}{2}R\), which applies at constant pressure.

Answer: D

The amount of heat added is 200 C_V.

At constant volume no work is done (\(W = P\Delta V = 0\)), so by the first law all the heat supplied goes into internal energy. The temperature rises by \(89 - (-11) = 100\) degrees, and with \(C_{V}\) the molar heat capacity at constant volume,

\[ Q = nC_{V}\Delta T = 2 \times C_{V} \times 100 = 200\,C_{V} \]

The internal energy also increases by \(200\,C_{V}\), not 100 \(C_{V}\), which forgets the 2 moles. No work is done on or by the gas, and the internal energy certainly changes, since the temperature rises.

Answer: C

The engine should absorb heat at 227°C.

The efficiency of an ideal (Carnot) engine is \(\eta = 1 - \frac{T_{2}}{T_{1}}\), with temperatures in kelvin. The sink is at 27 °C = 300 K.

\[ \begin{aligned} 0.4 &= 1 - \frac{300}{T_{1}} \\ T_{1} &= \frac{300}{0.6} = 500 \text{ K} = 227\,^\circ\text{C} \end{aligned} \]

500°C is the kelvin value written as Celsius, and 427°C comes from \(T_{1} = 700\) K, which would give an efficiency of about 57 %.

Answer: D

The heat required at constant volume is 217 J.

For one mole, \(C_p - C_v = R\). The extra heat at constant pressure is the work done in expansion, \(nR\Delta T\).

\[ \begin{aligned} Q_v &= Q_p - nR\Delta T \\ &= 300 - 1 \times 8.3 \times 10 = 217\ \text{J} \end{aligned} \]

83 J is the work done at constant pressure, and 300 J is the heat at constant pressure itself. 383 J would add the work instead of subtracting it.

Answer: B

The source temperature should be raised to 600 K.

The present efficiency is \(1 - \frac{300}{500} = 0.4\), that is 40%. Increasing it by 10% here means by 10 percentage points, to 50%.

\[ \begin{aligned} 1 - \frac{300}{T_1} &= 0.5 \\ T_1 &= 600\ \text{K} \end{aligned} \]

Reading '10%' as a relative increase (to 44%) gives about 536 K, which is not offered, so the percentage-point reading is intended. 550 K is a 10% rise in the source temperature itself.

Answer: D

The volume first decreases and then increases.

Water behaves unusually between 0 and 4 degrees Celsius. As ice-cold water is warmed, it contracts until about 4 degrees Celsius, where it has its greatest density and smallest volume, because the open, hydrogen-bonded structure left over from ice breaks down. Above 4 degrees it expands normally with heating, so from 4 to 10 degrees its volume increases.

'It shall increase' describes ordinary liquids and ignores the anomaly; 'It shall decrease' is true only up to 4 degrees; and 'first increase and then decrease' reverses the actual order.

Answer: D

The key's answer is Entropy remains constant.

In an adiabatic process no heat enters or leaves. For a reversible (quasi-static) adiabatic process, dS = dQ_rev/T = 0, so the entropy stays the same; this is why a reversible adiabatic process is called isentropic. The answer assumes that standard textbook case; a sudden free expansion is also adiabatic but increases entropy.

The other options are wrong. The gas does work while expanding with no heat supplied, so its internal energy and therefore its temperature fall, and the pressure falls too (PV^γ is constant).

Answer: B

For an irreversible process, \(\Delta S \gt 0\): the total entropy increases.

The second law of thermodynamics says that the total entropy of a system and its surroundings never decreases. It stays constant only in an ideal reversible process and increases in every real, irreversible one, such as heat flowing across a finite temperature difference or a gas expanding freely.

So ΔS = 0 describes a reversible process, ΔS below zero is impossible for the system and surroundings together, and 'None of these' is wrong.

Answer: D

The normal body temperature is about 310K.

Normal body temperature is about 37°C. To convert Celsius to kelvin, add 273:

\[ \begin{aligned} T &= 37 + 273 \\ &= 310\ \text{K} \end{aligned} \]

280 K, 290 K and 300 K correspond to about 7°C, 17°C and 27°C, all far below body temperature.

As a check, 37°C is 98.6°F, the familiar Fahrenheit figure for normal body temperature, and both describe the same 310 K.