RowQ
The Vault
RowQ
The Vault
CBSE Class 11 Physics · 10 questions · 24 marks
Heat is energy in transit, while temperature tells you which way that energy will flow. This chapter shows you how solids and liquids swell when heated, how to account for every joule using calorimetry, and what happens to that energy during melting and boiling when the thermometer refuses to move. You will also compare the three ways heat travels — conduction, convection and radiation — and see why a thermos flask must defeat all three.
A metal rod has a coefficient of linear expansion α. The coefficient of volume expansion of the same metal is:
Answer
3α — consider a cube of side L, whose volume is V = L³. On heating, each side becomes L(1 + αΔT), so the new volume is L³(1 + αΔT)³ ≈ L³(1 + 3αΔT), because αΔT is very small and the terms in (αΔT)² and higher can be dropped. Comparing this with V(1 + γΔT) gives γ = 3α. By the same argument the area expansion coefficient β = 2α.
Ice at 0 °C is added to water at 0 °C in an insulated flask. What happens?
Answer
Nothing happens, since there is no temperature difference to drive heat flow — heat flows only from a higher temperature to a lower one, and here both substances sit at exactly 0 °C. Melting the ice would need latent heat that nothing can supply, and freezing the water would need that latent heat to be carried away, which the insulation prevents. The mixture simply stays as it is.
Two rods of the same length and cross-section, one of thermal conductivity 2K and the other of conductivity K, are joined end to end. Their outer ends are held at 100 °C and 0 °C respectively, the better conductor being on the hot side. The steady-state temperature of the junction is:
Answer
66.7 °C — in the steady state the same heat current passes through both rods. So 2KA(100 − T)/L = KA(T − 0)/L. Cancelling A and L gives 2(100 − T) = T, hence 200 = 3T and T = 66.7 °C. The larger temperature drop always occurs across the poorer conductor, which is why the junction sits closer to the hot end here.
A blackbody at temperature T radiates power P. If its absolute temperature is doubled while its area stays the same, the power radiated becomes:
Answer
16P — by the Stefan-Boltzmann law P = σAT⁴, the emitted power depends on the fourth power of the absolute temperature. Doubling T multiplies the power by 2⁴ = 16. This steep dependence is why a filament that is only a few times hotter than a hotplate glows enormously more brightly, and why radiation dominates all other loss mechanisms at high temperatures.
Assertion (A): A steel ring that is slightly too small to fit over a steel shaft can be slipped on after it is heated strongly. Reason (R): On heating, the hole in the ring expands in the same proportion as the surrounding metal, so its diameter increases.
Answer
Both A and R are true and R is the correct explanation of A — a common misconception is that the metal expands inwards and closes the hole. In fact every linear dimension of the object, including the diameter of any hole in it, scales by the same factor (1 + αΔT). The hole behaves exactly like a disc of the same metal that has been removed. So heating enlarges the bore and the ring slides on; on cooling it grips the shaft tightly, which is how railway wheels are shrink-fitted onto their axles.
A brass measuring tape is correct at 20 °C. It is used on a hot afternoon when the temperature is 45 °C to measure a distance, and reads 60.000 m. What is the true distance? Take α_brass = 19×10⁻⁶ /°C.
Answer
On a hot day each division of the tape has expanded, so the tape reads less than the true length. ΔT = 45 − 20 = 25 °C. Each marked metre is actually 1 × (1 + αΔT) = 1 + (19×10⁻⁶ × 25) = 1 + 4.75×10⁻⁴ m long. True distance = reading × (1 + αΔT) = 60.000 × (1 + 4.75×10⁻⁴) = 60.000 + 60 × 4.75×10⁻⁴ = 60.000 + 0.0285 = 60.0285 m. So the true distance is about 60.029 m — the tape under-reads by roughly 2.9 cm.
A 0.15 kg block of an unknown alloy is heated to 150 °C and dropped into 0.30 kg of water at 22 °C held in a calorimeter of negligible heat capacity. The final steady temperature is 28 °C. Find the specific heat capacity of the alloy. Take c_water = 4200 J/kg·K.
Answer
By the principle of calorimetry, heat lost by the alloy = heat gained by the water. Heat gained by water: Q = m_w c_w ΔT_w = 0.30 × 4200 × (28 − 22) = 0.30 × 4200 × 6 = 7560 J. Heat lost by alloy: Q = m_a c_a ΔT_a = 0.15 × c_a × (150 − 28) = 0.15 × 122 × c_a = 18.3 c_a. Equating: 18.3 c_a = 7560 c_a = 7560/18.3 ≈ 413 J/kg·K. So the specific heat capacity of the alloy is about 4.1×10² J/kg·K, roughly a tenth that of water — which is why the small hot block raises the water's temperature by only 6 °C.
(a) Define the coefficient of linear expansion and derive the relation γ = 3α between the volume and linear expansion coefficients of an isotropic solid. (b) A steel bridge girder is 40 m long at 15 °C. Calculate the expansion gap that must be left at each end if the temperature can rise to 47 °C. Take α_steel = 1.2×10⁻⁵ /°C. (c) Explain what would happen if no gap were left, and estimate the thermal stress produced, given Y_steel = 2×10¹¹ Pa.
Answer
(a) Definition: the coefficient of linear expansion α is the fractional increase in length of a solid per unit rise in temperature, α = ΔL/(LΔT), with unit /°C or /K. Derivation of γ = 3α: take a cube of the material with side L at temperature T, so its volume is V = L³. Raise the temperature by ΔT. Every linear dimension grows by the same factor, so the new side is L' = L(1 + αΔT). The new volume is therefore V' = L'³ = L³(1 + αΔT)³. Expanding the cube of the bracket: (1 + αΔT)³ = 1 + 3αΔT + 3(αΔT)² + (αΔT)³. Since α is of the order of 10⁻⁵ and ΔT is at most a few hundred degrees, the quantity αΔT is much smaller than 1, so the squared and cubed terms are utterly negligible. Hence V' ≈ L³(1 + 3αΔT) = V(1 + 3αΔT). But by definition V' = V(1 + γΔT). Comparing the two expressions: γ = 3α. The same argument applied to a square face gives (1 + αΔT)² ≈ 1 + 2αΔT, so β = 2α. Thus α : β : γ = 1 : 2 : 3. (b) ΔT = 47 − 15 = 32 °C. Total expansion ΔL = αLΔT = 1.2×10⁻⁵ × 40 × 32 = 1.2×10⁻⁵ × 1280 = 1.536×10⁻² m = 15.36 mm. If the girder is free to expand at both ends, a gap of about 7.7 mm at each end is enough; if one end is fixed, the full 15.4 mm must be provided at the other. (c) With no gap, the girder cannot lengthen, so the supports exert compressive forces that squeeze it back by exactly the amount it would have expanded. The strain that has been prevented is strain = ΔL/L = αΔT = 1.2×10⁻⁵ × 32 = 3.84×10⁻⁴. The thermal stress is therefore stress = Y × strain = 2×10¹¹ × 3.84×10⁻⁴ = 7.68×10⁷ Pa. That is roughly 77 MPa of compression produced by nothing more than a warm afternoon, quite independent of the girder's length. Such stresses can buckle the deck sideways or crack the supporting piers, which is exactly why bridges carry toothed expansion joints and railway tracks are laid with small gaps or welded under controlled tension.
A copper vessel of mass 0.20 kg contains 0.50 kg of water, both at 30 °C. A 0.080 kg lump of ice at −10 °C is dropped in. Find the final temperature of the contents, assuming no heat is exchanged with the surroundings. Take c_copper = 390 J/kg·K, c_water = 4200 J/kg·K, c_ice = 2100 J/kg·K and L_fusion = 3.34×10⁵ J/kg. Explain clearly why the answer must be checked in stages.
Answer
Strategy: this kind of problem must be worked in stages because the ice may or may not melt completely. Compare the heat available from the warm contents with the heat the ice demands. Step 1 — heat available if everything cools to 0 °C. From the water: Q = 0.50 × 4200 × (30 − 0) = 63000 J. From the copper vessel: Q = 0.20 × 390 × (30 − 0) = 2340 J. Total heat available down to 0 °C = 63000 + 2340 = 65340 J. Step 2 — heat the ice needs. To warm the ice from −10 °C to 0 °C: Q = 0.080 × 2100 × 10 = 1680 J. To melt all of it at 0 °C: Q = mL = 0.080 × 3.34×10⁵ = 26720 J. Total heat needed by the ice = 1680 + 26720 = 28400 J. Step 3 — decide the outcome. Since 65340 J available is greater than 28400 J needed, all the ice warms up and melts completely, and there is still 65340 − 28400 = 36940 J of surplus that has not been used. The final temperature must therefore lie above 0 °C, and we solve for it. Step 4 — find the common final temperature T. Heat lost by the original water and vessel in cooling from 30 °C to T: Q_lost = (0.50 × 4200 + 0.20 × 390)(30 − T) = (2100 + 78)(30 − T) = 2178(30 − T). Heat gained by the ice: warming to 0 °C, melting, then the melt-water warming from 0 °C to T: Q_gained = 1680 + 26720 + (0.080 × 4200 × T) = 28400 + 336T. Setting Q_lost = Q_gained: 2178(30 − T) = 28400 + 336T 65340 − 2178T = 28400 + 336T 65340 − 28400 = 336T + 2178T 36940 = 2514T T = 36940/2514 ≈ 14.7 °C. Final answer: the contents settle at about 14.7 °C, and the vessel then holds 0.58 kg of water. Why the staged check matters: had the ice been much more massive, the heat needed would have exceeded 65340 J and the answer would instead have been 0 °C with a mixture of ice and water, the amount melted being found from the leftover heat. Writing one blind equation would then have produced a nonsensical negative temperature, so the comparison in Step 3 is an essential part of the method.
Read the following and answer the questions that follow: A food-storage crate has a wall of area 1.5 m² made of a composite panel. The inner layer is 0.02 m of foam with thermal conductivity 0.04 W/m·K, and the outer layer is 0.006 m of plywood with thermal conductivity 0.12 W/m·K. The inside is held at 4 °C while the outside surface is at 34 °C, and the flow has reached a steady state. (a) Calculate the thermal resistance of each layer and of the wall as a whole. (b) Find the steady rate at which heat leaks into the crate. (c) Find the temperature at the boundary between the foam and the plywood. (d) The designer proposes replacing the plywood with a metal sheet of the same thickness to make the crate stronger. Comment on the effect this would have on the heat leak, and state which layer is doing the real insulating work.
Answer
(a) Thermal resistance of a slab is R = L/(KA). Foam: R₁ = 0.02/(0.04 × 1.5) = 0.02/0.06 ≈ 0.333 K/W. Plywood: R₂ = 0.006/(0.12 × 1.5) = 0.006/0.18 ≈ 0.0333 K/W. The two layers carry the same heat current one after the other, so they are in series and their resistances simply add: R_total = 0.333 + 0.0333 ≈ 0.367 K/W. (b) The overall temperature difference is ΔT = 34 − 4 = 30 °C. H = ΔT/R_total = 30/0.367 ≈ 81.8 W. So about 82 J of heat enters the crate every second through this wall. (c) The same current H flows through the plywood alone, between the outer surface at 34 °C and the interface at temperature T. H = (34 − T)/R₂ 81.8 = (34 − T)/0.0333 34 − T = 81.8 × 0.0333 ≈ 2.72 T ≈ 31.3 °C. So the interface sits at about 31.3 °C. Notice that almost the whole 30 °C drop — some 27.3 °C of it — occurs across the foam. (d) A metal sheet would have a thermal conductivity of the order of 50 to 400 W/m·K, thousands of times larger than plywood's 0.12 W/m·K, so its resistance would be effectively zero and R_total would fall to about 0.333 K/W. The heat leak would rise only slightly, from about 82 W to about 90 W, an increase of roughly 10%. This is because in a series arrangement the largest resistance dominates, and here the foam contributes about 91% of the total. The foam is doing essentially all of the insulating work; the outer skin is structural, and swapping it for metal costs surprisingly little in thermal performance, though the metal would then feel cold to the touch and could collect condensation.
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