401. A \(0.20\,\text{kg}\) block of ice at \(-10^\circ\text{C}\) is heated to water at \(20^\circ\text{C}\). Take \(c_{\text{ice}}=2100\,\text{J kg}^{-1}\text{K}^{-1}\), \(L_f=3.36\times10^5\,\text{J kg}^{-1}\), and \(c_w=4200\,\text{J kg}^{-1}\text{K}^{-1}\). What total heat is required?
ⓐ. \(8.82\times10^4\,\text{J}\)
ⓑ. \(9.24\times10^4\,\text{J}\)
ⓒ. \(1.34\times10^5\,\text{J}\)
ⓓ. \(6.72\times10^4\,\text{J}\)
Correct Answer: \(8.82\times10^4\,\text{J}\)
Explanation: \( \textbf{Step 1: Warm ice from \(-10^\circ\text{C}\) to \(0^\circ\text{C}\).} \)
\[
Q_1=m c_{\text{ice}}\Delta T
\]
\[
Q_1=(0.20)(2100)(10)
\]
\[
Q_1=4200\,\text{J}
\]
\( \textbf{Step 2: Melt ice at \(0^\circ\text{C}\).} \)
\[
Q_2=mL_f
\]
\[
Q_2=(0.20)(3.36\times10^5)
\]
\[
Q_2=67200\,\text{J}
\]
\( \textbf{Step 3: Warm the resulting water from \(0^\circ\text{C}\) to \(20^\circ\text{C}\).} \)
\[
Q_3=m c_w\Delta T
\]
\[
Q_3=(0.20)(4200)(20)
\]
\[
Q_3=16800\,\text{J}
\]
Total heat required is
\[
Q=Q_1+Q_2+Q_3
\]
\[
Q=4200+67200+16800
\]
\[
Q=88200\,\text{J}=8.82\times10^4\,\text{J}
\]
All three stages are needed because the substance starts as ice below its melting point and ends as liquid water above it.
\( \textbf{Final answer:} \) \(8.82\times10^4\,\text{J}\).
402. A \(1.0\,\text{m}\) pendulum clock is correct at \(20^\circ\text{C}\). On a hot day, its pendulum length increases by \(0.04\%\). The time period changes approximately by:
ⓐ. \(0.08\%\) increase
ⓑ. \(0.04\%\) increase
ⓒ. \(0.04\%\) decrease
ⓓ. \(0.02\%\) increase
Correct Answer: \(0.02\%\) increase
Explanation: The time period of a simple pendulum is \[
T=2\pi\sqrt{\frac{l}{g}}
\]
For small changes, the fractional change in time period is half the fractional change in length: \[
\frac{\Delta T}{T}=\frac{1}{2}\frac{\Delta l}{l}
\]
The length increase is \(0.04\%\). Therefore, \[
\frac{\Delta T}{T}=\frac{1}{2}(0.04\%)=0.02\%
\]
The period increases because the pendulum becomes longer. A larger period means each swing takes slightly more time, so the clock tends to run slow. \( \textbf{Final answer:} \) The time period increases by about \(0.02\%\).
403. A bimetallic strip is made of metals \(P\) and \(Q\), with \(\alpha_P\gt\alpha_Q\). On cooling the strip below its original temperature, the strip bends so that:
ⓐ. metal \(Q\) disappears from the strip
ⓑ. metal \(P\) is on the inner side of the curve
ⓒ. both metals must remain straight because cooling prevents bending
ⓓ. metal \(P\) is on the outer side of the curve
Correct Answer: metal \(P\) is on the inner side of the curve
Explanation: On heating, the metal with larger \(\alpha\) tends to become longer and therefore lies on the outer side of the bend. On cooling, the same metal tends to contract more than the other metal. Since the strips are bonded, unequal contraction again causes bending. The metal that contracts more becomes shorter and lies on the inner side of the curve. Therefore, metal \(P\), with larger \(\alpha\), is on the inner side during cooling below the original temperature. The direction of bending reverses when the temperature change changes sign.
404. A metal rod of length \(2.0\,\text{m}\) and area \(5.0\times10^{-4}\,\text{m}^2\) is fixed between rigid supports. Given \(Y=2.0\times10^{11}\,\text{Pa}\), \(\alpha=1.0\times10^{-5}\,\text{K}^{-1}\), and a temperature fall of \(30\,\text{K}\), what is the magnitude of the thermal force developed?
ⓐ. \(3.0\times10^4\,\text{N}\)
ⓑ. \(6.0\times10^4\,\text{N}\)
ⓒ. \(1.0\times10^4\,\text{N}\)
ⓓ. \(2.0\times10^4\,\text{N}\)
Correct Answer: \(3.0\times10^4\,\text{N}\)
Explanation: \( \textbf{Given data:} \) \(A=5.0\times10^{-4}\,\text{m}^2\), \(Y=2.0\times10^{11}\,\text{Pa}\), \(\alpha=1.0\times10^{-5}\,\text{K}^{-1}\), and \(|\Delta T|=30\,\text{K}\). If expansion or contraction is fully prevented, thermal stress magnitude is \[
\text{stress}=Y\alpha|\Delta T|
\]
Substitute values: \[
\text{stress}=(2.0\times10^{11})(1.0\times10^{-5})(30)
\]
\[
\text{stress}=6.0\times10^7\,\text{Pa}
\]
Force is stress multiplied by area: \[
F=(6.0\times10^7)(5.0\times10^{-4})
\]
\[
F=3.0\times10^4\,\text{N}
\]
Cooling would produce tensile stress if contraction is prevented, but the question asks for magnitude. \( \textbf{Final answer:} \) The force is \(3.0\times10^4\,\text{N}\).
405. A composite wall has two layers in series with the same area. Layer \(P\) has \(L_P=0.10\,\text{m}\) and \(k_P=0.50\,\text{W m}^{-1}\text{K}^{-1}\). Layer \(Q\) has \(L_Q=0.20\,\text{m}\) and \(k_Q=1.0\,\text{W m}^{-1}\text{K}^{-1}\). Which statement about their temperature drops in steady state is correct?
ⓐ. The better conductor must have all the temperature drop
ⓑ. Layer \(Q\) has twice the temperature drop of layer \(P\)
ⓒ. The two layers have equal temperature drops
ⓓ. Layer \(P\) has twice the temperature drop of layer \(Q\)
Correct Answer: The two layers have equal temperature drops
Explanation: Thermal resistance of a layer is \(R=\frac{L}{kA}\). Since both layers have the same area \(A\), compare \(\frac{L}{k}\). For layer \(P\), \(\frac{L_P}{k_P}=\frac{0.10}{0.50}=0.20\). For layer \(Q\), \(\frac{L_Q}{k_Q}=\frac{0.20}{1.0}=0.20\). Their thermal resistances are equal. In series, the same heat current flows through both layers, and temperature drop is proportional to resistance. Equal resistances therefore produce equal temperature drops.
406. Two rods are connected in parallel between the same hot and cold reservoirs. Rod \(P\) and rod \(Q\) have the same length and area, but \(k_P=3k_Q\). The ratio of heat currents \(H_P:H_Q\) is:
ⓐ. \(1:3\)
ⓑ. \(3:1\)
ⓒ. \(9:1\)
ⓓ. \(1:1\)
Correct Answer: \(3:1\)
Explanation: For each rod in parallel, the temperature difference across the rod is the same. The rods also have the same length \(L\) and area \(A\). Heat current through a rod is \[
H=\frac{kA\Delta T}{L}
\]
With \(A\), \(\Delta T\), and \(L\) common, \(H\propto k\). Since \(k_P=3k_Q\), \[
H_P:H_Q=3:1
\]
The currents are not forced to be equal in parallel; equality of heat current is a series condition. \( \textbf{Final answer:} \) \(H_P:H_Q=3:1\).
407. A slab has one face at \(100^\circ\text{C}\) and the other at \(40^\circ\text{C}\). Its thickness is \(12\,\text{cm}\), and the steady temperature variation is linear. What is the temperature at a point \(3\,\text{cm}\) from the hot face?
ⓐ. \(80^\circ\text{C}\)
ⓑ. \(85^\circ\text{C}\)
ⓒ. \(55^\circ\text{C}\)
ⓓ. \(70^\circ\text{C}\)
Correct Answer: \(85^\circ\text{C}\)
Explanation: \( \textbf{Surface temperatures:} \) Hot face \(=100^\circ\text{C}\), cold face \(=40^\circ\text{C}\). Total temperature drop is \[
100-40=60^\circ\text{C}
\]
The slab thickness is \(12\,\text{cm}\). A point \(3\,\text{cm}\) from the hot face is one-fourth of the way across the slab: \[
\frac{3}{12}=\frac{1}{4}
\]
Temperature drop over this distance is \[
\frac{1}{4}\times60=15^\circ\text{C}
\]
So the temperature at that point is \[
100-15=85^\circ\text{C}
\]
The linear profile is valid for steady conduction through a uniform slab with constant \(k\). \( \textbf{Final answer:} \) The temperature is \(85^\circ\text{C}\).
408. A rod conducts heat steadily at \(80\,\text{W}\). The heat arriving at a water bath is used only to raise the temperature of \(0.50\,\text{kg}\) water. How long will it take to raise the water temperature by \(12\,\text{K}\), taking \(c_w=4200\,\text{J kg}^{-1}\text{K}^{-1}\)?
ⓐ. \(630\,\text{s}\)
ⓑ. \(315\,\text{s}\)
ⓒ. \(105\,\text{s}\)
ⓓ. \(210\,\text{s}\)
Correct Answer: \(315\,\text{s}\)
Explanation: \( \textbf{Heat needed by water:} \) \[
Q=mc_w\Delta T
\]
\[
Q=(0.50)(4200)(12)
\]
\[
Q=25200\,\text{J}
\]
The steady heat current is \(H=80\,\text{W}=80\,\text{J s}^{-1}\). Use \[
H=\frac{Q}{t}
\]
So, \[
t=\frac{Q}{H}
\]
\[
t=\frac{25200}{80}=315\,\text{s}
\]
The problem combines conduction rate with heat capacity of the receiving water. \( \textbf{Final answer:} \) The time required is \(315\,\text{s}\).
409. A \(100\,\text{W}\) heater is used to warm \(0.50\,\text{kg}\) of water from \(20^\circ\text{C}\) to \(40^\circ\text{C}\). If only \(70\%\) of the heater energy reaches the water, how long does the heating take? Take \(c_w=4200\,\text{J kg}^{-1}\text{K}^{-1}\).
ⓐ. \(600\,\text{s}\)
ⓑ. \(300\,\text{s}\)
ⓒ. \(840\,\text{s}\)
ⓓ. \(420\,\text{s}\)
Correct Answer: \(600\,\text{s}\)
Explanation: Heat needed by water is \[
Q=mc_w\Delta T
\]
\[
Q=(0.50)(4200)(40-20)
\]
\[
Q=(0.50)(4200)(20)=42000\,\text{J}
\]
Only \(70\%\) of heater power reaches the water. Effective power is \[
P_{\text{eff}}=0.70(100)=70\,\text{W}
\]
Time required is \[
t=\frac{Q}{P_{\text{eff}}}
\]
\[
t=\frac{42000}{70}=600\,\text{s}
\]
Using the full \(100\,\text{W}\) would ignore the stated heat loss. \( \textbf{Final answer:} \) The heating takes \(600\,\text{s}\).
410. A graph of temperature against heat supplied for a pure substance has a steep rising segment for the solid, a horizontal melting segment, a less steep rising segment for the liquid, and a horizontal boiling segment. If the same mass is present throughout, the less steep liquid segment means:
ⓐ. the liquid has larger specific heat capacity than the solid
ⓑ. the liquid is changing state throughout the rising segment
ⓒ. no heat is being supplied to the liquid
ⓓ. the liquid has smaller specific heat capacity than the solid
Correct Answer: the liquid has larger specific heat capacity than the solid
Explanation: On a graph of temperature \(T\) against heat supplied \(Q\), the slope of a warming segment is \(\frac{\Delta T}{\Delta Q}\). From \(Q=mc\Delta T\), we get \(\frac{\Delta T}{\Delta Q}=\frac{1}{mc}\). For the same mass \(m\), a smaller slope means a larger specific heat capacity \(c\). The less steep liquid segment therefore shows that the liquid needs more heat for the same temperature rise than the solid. Horizontal parts represent phase changes, but the rising liquid segment represents warming within the liquid state. The slope comparison uses heat capacity, not latent heat.
411. In a heating curve, the boiling plateau is longer than the melting plateau for the same substance and same heating rate. What does this indicate?
ⓐ. \(c_{\text{liquid}}=0\)
ⓑ. \(L_v\gt L_f\)
ⓒ. boiling happens without energy
ⓓ. \(L_f\gt L_v\)
Correct Answer: \(L_v\gt L_f\)
Explanation: A plateau on a heating curve represents a phase change at constant temperature. The duration of the plateau depends on the heat needed for that phase change. For the same mass and same heating rate, longer time means larger heat input. Since phase-change heat is \(Q=mL\), the longer boiling plateau indicates a larger latent heat. Therefore, the latent heat of vaporisation \(L_v\) is greater than the latent heat of fusion \(L_f\). The temperature being constant during the plateau does not mean no energy is supplied.
412. A \(0.20\,\text{kg}\) sample of water at \(100^\circ\text{C}\) is mixed with \(0.010\,\text{kg}\) steam at \(100^\circ\text{C}\) in an insulated vessel. The steam condenses completely, and the final temperature is below \(100^\circ\text{C}\). Take \(L_v=2.26\times10^6\,\text{J kg}^{-1}\) and \(c_w=4200\,\text{J kg}^{-1}\text{K}^{-1}\). What is the heat released by steam only during condensation?
ⓐ. \(2.26\times10^3\,\text{J}\)
ⓑ. \(2.26\times10^5\,\text{J}\)
ⓒ. \(2.26\times10^4\,\text{J}\)
ⓓ. \(4.52\times10^4\,\text{J}\)
Correct Answer: \(2.26\times10^4\,\text{J}\)
Explanation: \( \textbf{Mass of steam:} \) \(m_s=0.010\,\text{kg}\). The question asks only for heat released during condensation at \(100^\circ\text{C}\). Use \[
Q=m_sL_v
\]
Substitute values: \[
Q=(0.010)(2.26\times10^6)
\]
\[
Q=2.26\times10^4\,\text{J}
\]
If the condensed water later cools below \(100^\circ\text{C}\), that releases additional \(mc\Delta T\) heat. The requested quantity isolates only the latent heat released in condensation. \( \textbf{Final answer:} \) The condensation heat is \(2.26\times10^4\,\text{J}\).
413. A body with heat capacity \(C\) cools by radiation in surroundings at temperature \(T_s\). If its temperature is \(T\), the instantaneous rate of fall of temperature due to net radiation is proportional to:
ⓐ. \(\frac{T^4-T_s^4}{C}\)
ⓑ. \(T_s^4-T^4\) when \(T\gt T_s\)
ⓒ. \(C(T^4-T_s^4)\)
ⓓ. \(\frac{C}{T^4-T_s^4}\)
Correct Answer: \(\frac{T^4-T_s^4}{C}\)
Explanation: Net radiative power loss is proportional to \(T^4-T_s^4\) when \(T\gt T_s\). Temperature fall is related to heat loss by \(Q=C\Delta T\). A larger heat capacity means more heat must be removed for the same fall in temperature. Therefore, for the same radiative power loss, the rate of temperature fall is smaller when \(C\) is larger. Symbolically, \(\frac{dT}{dt}\) in magnitude is proportional to \(\frac{T^4-T_s^4}{C}\). This combines the radiation law with the heat-capacity idea.
414. Two identical black bodies have absolute temperatures \(T_1\) and \(T_2\). If the wavelength of maximum emission of body \(1\) is three times that of body \(2\), then \(T_1:T_2\) is:
ⓐ. \(1:9\)
ⓑ. \(9:1\)
ⓒ. \(3:1\)
ⓓ. \(1:3\)
Correct Answer: \(1:3\)
Explanation: Wien’s displacement law states that the wavelength of maximum emission is inversely proportional to absolute temperature:
\[
\lambda_mT=b
\]
For body \(1\) and body \(2\), the same constant \(b\) applies because both are black bodies.
Given
\[
\lambda_{m,1}=3\lambda_{m,2}
\]
Using \(T\propto \frac{1}{\lambda_m}\),
\[
\frac{T_1}{T_2}=\frac{\lambda_{m,2}}{\lambda_{m,1}}
\]
Substitute the wavelength relation:
\[
\frac{T_1}{T_2}=\frac{\lambda_{m,2}}{3\lambda_{m,2}}
\]
\[
\frac{T_1}{T_2}=\frac{1}{3}
\]
Therefore,
\[
T_1:T_2=1:3
\]
The body with the longer peak wavelength is cooler, so body \(1\) has the lower absolute temperature. \( \textbf{Final answer:} \) \(T_1:T_2=1:3\).
415. A black body and a grey body have the same area and temperature. The grey body has emissivity \(e=0.60\). If the black body emits \(500\,\text{W}\), the grey body emits:
ⓐ. \(500\,\text{W}\)
ⓑ. \(200\,\text{W}\)
ⓒ. \(300\,\text{W}\)
ⓓ. \(833\,\text{W}\)
Correct Answer: \(300\,\text{W}\)
Explanation: For the same area and temperature, radiation power from a real body is \(e\) times the black-body power. Here, \(e=0.60\). Therefore, \[
P_{\text{grey}}=eP_{\text{black}}
\]
\[
P_{\text{grey}}=(0.60)(500)
\]
\[
P_{\text{grey}}=300\,\text{W}
\]
The grey body emits less than the black body because its emissivity is less than \(1\). Emissivity changes the radiated power without changing the area or temperature in this comparison. \( \textbf{Final answer:} \) The grey body emits \(300\,\text{W}\).
416. A body has temperature \(310\,\text{K}\), and the surroundings are at \(300\,\text{K}\). Another identical body has temperature \(330\,\text{K}\) in the same surroundings. Using Newton’s law of cooling, the ratio of initial cooling rates of the two bodies is:
ⓐ. \(1:1\)
ⓑ. \(1:3\)
ⓒ. \(1:2\)
ⓓ. \(31:33\)
Correct Answer: \(1:3\)
Explanation: Newton’s law of cooling states that cooling rate is proportional to temperature excess over surroundings. For the first body, the excess temperature is \[
310-300=10\,\text{K}
\]
For the second body, the excess temperature is \[
330-300=30\,\text{K}
\]
Thus the ratio of cooling rates is \[
10:30=1:3
\]
The absolute temperatures themselves are not compared directly; the excess temperatures are compared. This is why \(310:330\) is not the relevant ratio. \( \textbf{Final answer:} \) The ratio is \(1:3\).
417. A cup of tea cools from \(80^\circ\text{C}\) to \(70^\circ\text{C}\) in a room at \(30^\circ\text{C}\). Later, it cools from \(50^\circ\text{C}\) to \(40^\circ\text{C}\). Which interval should take longer according to Newton’s law of cooling?
ⓐ. both intervals must take equal time
ⓑ. \(50^\circ\text{C}\) to \(40^\circ\text{C}\)
ⓒ. \(80^\circ\text{C}\) to \(70^\circ\text{C}\)
ⓓ. neither interval occurs because cooling stops above room temperature
Correct Answer: \(50^\circ\text{C}\) to \(40^\circ\text{C}\)
Explanation: Newton’s law of cooling says that the cooling rate is larger when the temperature excess over surroundings is larger. During \(80^\circ\text{C}\) to \(70^\circ\text{C}\), the tea is far above the room temperature \(30^\circ\text{C}\). During \(50^\circ\text{C}\) to \(40^\circ\text{C}\), the excess temperature is smaller. A smaller cooling rate means the same \(10\,\text{K}\) temperature fall takes more time. The tea cools more slowly as it approaches room temperature. The later interval therefore lasts longer.
418. A thermos flask has a vacuum between its double walls and silvered surfaces. If the vacuum is lost but the silvering remains, which heat-transfer mode increases most directly between the walls?
ⓐ. latent heat transfer between walls
ⓑ. conduction and convection
ⓒ. only radiation
ⓓ. nuclear heating
Correct Answer: conduction and convection
Explanation: A vacuum reduces heat transfer by removing most matter between the walls. If the vacuum is lost, gas enters the space. The gas can conduct heat through molecular collisions and can also set up convection currents if temperature differences exist. The silvered surfaces still reduce radiation, but they do not prevent gas conduction and convection. Latent heat transfer would require a phase change, which is not the main design issue here. Losing the vacuum mainly weakens insulation against conduction and convection.
419. A wall consists of insulation and brick in series. In steady state, the insulation has a larger temperature drop than the brick. The best conclusion is:
ⓐ. the brick carries no heat current
ⓑ. the insulation must have higher thermal conductivity
ⓒ. the insulation carries larger heat current than the brick
ⓓ. the insulation has larger thermal resistance
Correct Answer: the insulation has larger thermal resistance
Explanation: In a series wall at steady state, the same heat current passes through every layer. Temperature drop across a layer is \(H R\), where \(R\) is thermal resistance. If the insulation has a larger temperature drop while carrying the same heat current, its thermal resistance is larger. A larger resistance usually comes from lower thermal conductivity, larger thickness, or both. The larger drop does not mean a larger heat current in that layer. Temperature drop identifies the layer that offers greater opposition to heat conduction.
420. A graph of \(\ln(T-T_s)\) against time \(t\) for a body obeying Newton’s law of cooling is expected to be:
ⓐ. a circle
ⓑ. a straight line with positive slope
ⓒ. a straight line with negative slope
ⓓ. a horizontal line at all temperatures
Correct Answer: a straight line with negative slope
Explanation: Newton’s law gives the excess temperature as \(\theta=\theta_0e^{-Kt}\), where \(\theta=T-T_s\). Taking natural logarithm gives \[
\ln\theta=\ln\theta_0-Kt
\]
This has the form \(y=c-mt\). Therefore, a graph of \(\ln(T-T_s)\) against time is a straight line with slope \(-K\). The negative slope shows that the temperature excess decreases with time. A direct \(T\)-versus-\(t\) graph is curved, but the logarithmic plot becomes linear.
