301. A conduction experiment uses the same slab but reverses the two face temperatures. Initially, face \(P\) is at \(80^\circ\text{C}\) and face \(Q\) is at \(30^\circ\text{C}\). Later, face \(P\) is at \(30^\circ\text{C}\) and face \(Q\) is at \(80^\circ\text{C}\). What changes?
ⓐ. Thermal conductivity changes sign
ⓑ. same magnitude, but direction reverses
ⓒ. Magnitude becomes zero because the same temperatures are used
ⓓ. Heat current remains in the same direction always
Correct Answer: same magnitude, but direction reverses
Explanation: The magnitude of steady conductive heat current is proportional to the temperature difference. In both cases, the temperature difference has magnitude \(50\,\text{K}\). Therefore, the magnitude of heat current remains the same if the slab geometry and material are unchanged. However, heat always flows from the hotter face to the colder face. Reversing the hot and cold faces reverses the direction of heat flow. Thermal conductivity \(k\) is a material property and does not become negative because the labels are interchanged.
302. A composite wall is made of two layers in series. Layer \(P\) and layer \(Q\) have the same area and the same thickness, but \(k_P=2k_Q\). In steady state, what is true about the heat current through the two layers?
ⓐ. Heat current through \(Q\) is twice that through \(P\)
ⓑ. Heat current is the same through both layers
ⓒ. Heat current is zero in the better conductor
ⓓ. Heat current through \(P\) is twice that through \(Q\)
Correct Answer: Heat current is the same through both layers
Explanation: In steady state, heat cannot accumulate at the junction of the two layers. The heat current entering layer \(P\) must be the same as the heat current leaving layer \(Q\). If the heat currents were different, the temperature of the junction would keep changing with time. The different thermal conductivities affect how the total temperature drop is shared between the layers. The poorer conductor has the larger temperature drop for the same heat current. Equality of heat current is a steady-state condition for layers in series.
303. Two conducting slabs \(P\) and \(Q\) are joined in series and have the same area. Their thermal resistances are \(R_P=\frac{L_P}{k_PA}\) and \(R_Q=\frac{L_Q}{k_QA}\). The equivalent thermal resistance is:
ⓐ. \(R_P+R_Q\)
ⓑ. \(R_P-R_Q\)
ⓒ. \(\frac{1}{R_P+R_Q}\)
ⓓ. \(\frac{R_PR_Q}{R_P+R_Q}\)
Correct Answer: \(R_P+R_Q\)
Explanation: Thermal resistance for conduction is defined as \(R=\frac{L}{kA}\). For slabs in series, the same heat current passes successively through each slab. The temperature drops across the slabs add up to the total temperature difference. Since each temperature drop is \(HR\), the total drop is \(H(R_P+R_Q)\). Therefore, the equivalent resistance is \(R_P+R_Q\). This is similar in form to electrical resistances in series, but the physical flow here is heat.
304. A slab has thermal resistance \(4\,\text{K W}^{-1}\). If a temperature difference of \(40\,\text{K}\) is maintained across it, the steady heat current is:
ⓐ. \(5\,\text{W}\)
ⓑ. \(40\,\text{W}\)
ⓒ. \(160\,\text{W}\)
ⓓ. \(10\,\text{W}\)
Correct Answer: \(10\,\text{W}\)
Explanation: \( \textbf{Given data:} \) Thermal resistance \(R=4\,\text{K W}^{-1}\), and temperature difference \(\Delta T=40\,\text{K}\).
Thermal resistance is related to heat current by
\[
R=\frac{\Delta T}{H}
\]
Rearrange for \(H\):
\[
H=\frac{\Delta T}{R}
\]
Substitute:
\[
H=\frac{40}{4}
\]
\[
H=10\,\text{W}
\]
The unit check works because \(\text{K}\) divided by \(\text{K W}^{-1}\) gives \(\text{W}\).
\( \textbf{Final answer:} \) The heat current is \(10\,\text{W}\).
305. Study the table and identify the row that correctly describes the effect of changing a slab parameter in steady conduction.
| Row | Change made | Effect on heat current \(H\), other factors fixed |
|---|
| P | Area \(A\) is doubled | \(H\) doubles |
| Q | Thickness \(L\) is doubled | \(H\) doubles |
| R | Temperature difference \(\Delta T\) is doubled | \(H\) becomes half |
| S | Thermal conductivity \(k\) is doubled | \(H\) becomes zero |
ⓐ. Row S only
ⓑ. Row R only
ⓒ. Row Q only
ⓓ. Row P only
Correct Answer: Row P only
Explanation: The conduction relation is \(H=\frac{kA\Delta T}{L}\). It shows that \(H\) is directly proportional to \(A\), \(k\), and \(\Delta T\). It is inversely proportional to \(L\). Therefore, doubling area doubles the heat current, so row P is correct. Doubling thickness halves the heat current, not doubles it. Doubling temperature difference doubles the heat current, and doubling thermal conductivity also doubles it. The row must match both the changed quantity and its correct proportional effect.
306. A room has a window made of glass of thickness \(4\,\text{mm}\). Replacing it with glass of thickness \(8\,\text{mm}\), while keeping the same area and temperature difference, reduces conductive heat loss because:
ⓐ. thermal resistance increases
ⓑ. temperature difference changes into heat capacity
ⓒ. thermal conductivity becomes infinite
ⓓ. area becomes zero
Correct Answer: thermal resistance increases
Explanation: The thermal resistance of a slab is \(R=\frac{L}{kA}\). If the thickness \(L\) is doubled while \(k\) and \(A\) remain the same, the thermal resistance doubles. Since heat current is \(H=\frac{\Delta T}{R}\), a larger resistance gives a smaller heat current for the same temperature difference. The glass does not become a perfect conductor. The area is unchanged in the stated comparison. Thicker insulating layers reduce heat transfer by increasing the conduction path.
307. A copper rod and a wooden rod have the same length and area and are kept between the same two temperatures. The copper rod conducts more heat per second mainly because:
ⓐ. copper has smaller area than the wooden rod
ⓑ. copper has larger thermal conductivity
ⓒ. heat flows from cold to hot in copper
ⓓ. wood has no temperature
Correct Answer: copper has larger thermal conductivity
Explanation: Thermal conductivity \(k\) measures how well a material conducts heat. With the same length, area, and temperature difference, the heat current is proportional to \(k\). Copper has a much larger thermal conductivity than wood, so it carries heat faster. Wood is a poor conductor and is often used where insulation is useful. The direction of heat flow is still from the hotter end to the colder end. The difference between the rods comes from the material property, not from the geometry in this comparison.
308. A composite slab has two layers of equal thickness and equal area. Layer \(P\) has thermal conductivity \(4k\), and layer \(Q\) has thermal conductivity \(k\). If the total temperature difference across the composite slab is \(50\,\text{K}\), what is the temperature drop across layer \(Q\) in steady state?
ⓐ. \(40\,\text{K}\)
ⓑ. \(50\,\text{K}\)
ⓒ. \(10\,\text{K}\)
ⓓ. \(20\,\text{K}\)
Correct Answer: \(40\,\text{K}\)
Explanation: \( \textbf{Layer condition:} \) The two layers have equal thickness \(L\) and equal area \(A\).
Thermal resistance is
\[
R=\frac{L}{kA}
\]
For layer \(P\),
\[
R_P=\frac{L}{(4k)A}=\frac{1}{4}\frac{L}{kA}
\]
For layer \(Q\),
\[
R_Q=\frac{L}{kA}
\]
So,
\[
R_P:R_Q=1:4
\]
In series, the same heat current passes through both layers, so temperature drops are proportional to thermal resistances.
The total temperature difference is \(50\,\text{K}\).
The drop across \(Q\) is
\[
\Delta T_Q=\frac{4}{1+4}\times50
\]
\[
\Delta T_Q=40\,\text{K}
\]
The poorer conductor gets the larger temperature drop in steady conduction.
\( \textbf{Final answer:} \) The temperature drop across layer \(Q\) is \(40\,\text{K}\).
309. A cooking pot is heated from below. Water near the bottom becomes warmer and rises, while cooler water moves downward. This circulation is mainly:
ⓐ. conduction through vacuum
ⓑ. radiation alone
ⓒ. convection
ⓓ. sublimation
Correct Answer: convection
Explanation: In water, heat transfer can occur through bulk motion of the liquid. Water near the bottom receives heat, expands slightly, and becomes less dense. The warmer water rises, and cooler water moves downward to replace it. This circulating motion transfers thermal energy through the liquid. Conduction also occurs locally, but the large-scale circulation is convection. Sublimation is a change from solid directly to vapour, which is not the process described.
310. A room heater warms the air near it. The air rises and cooler air moves in near the floor. The repeated motion forms:
ⓐ. a conduction path through a solid
ⓑ. a convection current
ⓒ. a vacuum radiation shield
ⓓ. a latent heat plateau
Correct Answer: a convection current
Explanation: A convection current is a circulating flow set up in a fluid due to temperature and density differences. Air near the heater becomes warmer, expands, and becomes less dense. It rises, while cooler and denser air moves in to replace it. This repeated movement distributes heat through the room. The air itself is moving in bulk, so the process is convection rather than pure conduction. A solid conduction path is not the main feature of this room-heating example.
311. Radiation differs from conduction and convection because radiation:
ⓐ. cannot occur between bodies at different temperatures
ⓑ. requires only liquids
ⓒ. requires only solids
ⓓ. does not require a material medium
Correct Answer: does not require a material medium
Explanation: Radiation transfers thermal energy by electromagnetic waves. Because electromagnetic waves can travel through vacuum, radiation does not require matter between the source and receiver. Conduction requires material contact or particle-level energy transfer through matter. Convection requires bulk motion of a fluid. Radiation can also occur through transparent media such as air, but its special feature is that it can cross empty space. Heat from the Sun reaching Earth is the most familiar example.
312. A dull black surface and a shiny polished surface are kept at the same high temperature. The dull black surface generally:
ⓐ. emits no radiation at all
ⓑ. emits thermal radiation more effectively
ⓒ. conducts heat only through vacuum
ⓓ. must have lower temperature because it is black
Correct Answer: emits thermal radiation more effectively
Explanation: Surface nature affects thermal radiation. A dull black surface is generally a good emitter and good absorber of radiation. A shiny polished surface is a poor emitter and poor absorber compared with a dull black surface. If both surfaces are at the same temperature and have similar area, the black surface emits more thermal radiation. The colour or finish does not by itself mean a lower temperature. Radiation depends on temperature, area, and surface character.
313. A thermos flask reduces heat transfer by using several design features. Which matching is most suitable?
| Feature | Main heat-transfer reduction |
|---|
| P. Vacuum between double walls | 1. Reduces conduction and convection |
| Q. Silvered inner surfaces | 2. Reduces radiation |
| R. Poor-conducting stopper | 3. Reduces conduction through the mouth |
ⓐ. P-2, Q-1, R-3
ⓑ. P-3, Q-2, R-1
ⓒ. P-1, Q-3, R-2
ⓓ. P-1, Q-2, R-3
Correct Answer: P-1, Q-2, R-3
Explanation: A vacuum between the double walls greatly reduces conduction and prevents convection because both require matter. Silvered surfaces reduce radiation because polished shiny surfaces are poor emitters and absorbers. A poor-conducting stopper reduces heat transfer through the mouth of the flask by conduction. No single feature removes all modes of heat transfer completely. The flask design works by reducing each mode as much as practical. The matching must connect each feature with the heat-transfer mechanism it most directly limits.
314. Assertion: A blackened vessel cools faster than a polished vessel of the same size at the same temperature, under similar surroundings.
Reason: A dull black surface is a better emitter of thermal radiation than a polished surface.
ⓐ. Assertion is true, but Reason is false
ⓑ. Assertion is false, but Reason is true
ⓒ. Both Assertion and Reason are true, and Reason explains Assertion
ⓓ. Both Assertion and Reason are true, but Reason does not explain Assertion
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: A hot body loses heat partly by radiation. A dull black surface emits radiation more effectively than a shiny polished surface at the same temperature. Therefore, a blackened vessel can lose heat faster by radiation under similar conditions. The Assertion is true when other factors such as size, temperature, and surroundings are comparable. The Reason is also true and gives the physical basis for the faster cooling. Surface finish matters because radiation is strongly affected by emissive power.
315. A white or shiny outer surface is often preferred for containers meant to reduce heating by sunlight because such a surface:
ⓐ. absorbs all radiation perfectly
ⓑ. reflects more radiation and absorbs less
ⓒ. stops conduction inside the container completely
ⓓ. makes convection impossible in air
Correct Answer: reflects more radiation and absorbs less
Explanation: Sunlight transfers energy mainly by radiation. A shiny or white surface reflects a larger fraction of incident radiation than a dull black surface. It therefore absorbs less radiant energy and heats up less under similar exposure. This does not completely stop conduction or convection once temperature differences exist. The surface mainly controls absorption of radiation. The choice of outer finish is a practical use of the difference between good and poor absorbers.
316. In a comparison of heat-transfer modes, the statement that best separates the three modes is:
ⓐ. conduction, convection, and radiation all require a solid medium
ⓑ. conduction occurs only through vacuum, while radiation occurs only in liquids
ⓒ. conduction uses contact, convection fluid motion, radiation no medium
ⓓ. convection and radiation are both changes of state at constant temperature
Correct Answer: conduction uses contact, convection fluid motion, radiation no medium
Explanation: Conduction transfers heat through matter without requiring the material as a whole to move. Convection transfers heat by the bulk motion of a fluid, so it occurs in liquids and gases. Radiation transfers heat by electromagnetic waves and can occur through vacuum. These differences help identify the dominant mode in examples such as a hot spoon, boiling water, and sunlight reaching Earth. The three modes can occur together in real situations, but their mechanisms are different. A correct comparison must mention both the medium requirement and the way energy is carried.
317. In steady conduction through a uniform slab, the temperature inside the slab changes linearly from the hot face to the cold face. This happens when:
ⓐ. heat transfer is taking place only by radiation through vacuum
ⓑ. the slab is melting throughout its thickness
ⓒ. the two faces are at the same temperature
ⓓ. constant \(k\) and no heat stored in the slab
Correct Answer: constant \(k\) and no heat stored in the slab
Explanation: In steady conduction, the temperature at each point inside the slab remains constant with time. Heat enters one face and leaves the other face at the same rate, so there is no continuous storage of heat inside the slab. For a uniform slab with constant thermal conductivity, the temperature falls uniformly with distance from the hot face to the cold face. This produces a straight-line temperature-distance graph. If both faces were at the same temperature, there would be no net heat current. The linear temperature profile is a steady-state conduction result, not a radiation-through-vacuum effect.
318. A uniform wall has its inner surface at \(60^\circ\text{C}\) and outer surface at \(20^\circ\text{C}\). If the temperature variation across the wall is linear, the temperature halfway through the wall is:
ⓐ. \(40^\circ\text{C}\)
ⓑ. \(80^\circ\text{C}\)
ⓒ. \(60^\circ\text{C}\)
ⓓ. \(20^\circ\text{C}\)
Correct Answer: \(40^\circ\text{C}\)
Explanation: \( \textbf{Surface temperatures:} \) Hot face \(=60^\circ\text{C}\), cold face \(=20^\circ\text{C}\).
For a uniform slab in steady state, the temperature changes linearly with distance.
The total temperature drop is
\[
60^\circ\text{C}-20^\circ\text{C}=40^\circ\text{C}
\]
Halfway through the wall, half of this drop has occurred.
Half of \(40^\circ\text{C}\) is
\[
20^\circ\text{C}
\]
So the midpoint temperature is
\[
60^\circ\text{C}-20^\circ\text{C}=40^\circ\text{C}
\]
This midpoint result uses the linear temperature profile of a homogeneous wall.
\( \textbf{Final answer:} \) The temperature halfway through the wall is \(40^\circ\text{C}\).
319. Two layers of equal thickness and equal area are joined in series. Layer \(P\) has smaller thermal conductivity than layer \(Q\). In steady conduction, the temperature-distance graph is steeper in:
ⓐ. both layers equally always
ⓑ. neither layer because temperature cannot vary in solids
ⓒ. layer \(Q\)
ⓓ. layer \(P\)
Correct Answer: layer \(P\)
Explanation: In steady state, the same heat current passes through both layers. The temperature drop across a layer is related to its thermal resistance. For equal thickness and area, the layer with smaller thermal conductivity has larger thermal resistance. A larger temperature drop over the same thickness means a steeper temperature-distance graph. Therefore, the poorer conductor has the steeper temperature gradient. This is why insulating layers show a larger temperature drop for the same heat current.
320. A composite slab has two layers in series with thermal resistances \(2\,\text{K W}^{-1}\) and \(3\,\text{K W}^{-1}\). A temperature difference of \(50\,\text{K}\) is maintained across the whole slab. The steady heat current is:
ⓐ. \(5\,\text{W}\)
ⓑ. \(10\,\text{W}\)
ⓒ. \(50\,\text{W}\)
ⓓ. \(25\,\text{W}\)
Correct Answer: \(10\,\text{W}\)
Explanation: \( \textbf{Thermal resistances in series:} \) \(R_1=2\,\text{K W}^{-1}\), \(R_2=3\,\text{K W}^{-1}\).
For layers in series, the equivalent thermal resistance is
\[
R_{\text{eq}}=R_1+R_2
\]
\[
R_{\text{eq}}=2+3=5\,\text{K W}^{-1}
\]
The heat current is related to temperature difference by
\[
H=\frac{\Delta T}{R_{\text{eq}}}
\]
Substitute \(\Delta T=50\,\text{K}\):
\[
H=\frac{50}{5}
\]
\[
H=10\,\text{W}
\]
The same heat current flows through both layers in the steady state.
\( \textbf{Final answer:} \) The heat current is \(10\,\text{W}\).
