Thermal Properties of Matter
ICSE · Class 11 · Physics
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A steel rod of length 1 m and cross-sectional area 1 cm² is heated from 0°C to 200°C. If the rod is not allowed to expand (rigidly clamped at both ends), what is the compressive stress developed in the rod? (Given: α for steel = 1.2 × 10⁻⁵ /°C, Young's modulus Y = 2 × 10¹¹ Pa)
A platinum resistance thermometer records a resistance of 5 Ω at 0°C and 7 Ω at 100°C. When immersed in a hot liquid, it records 6.5 Ω. What is the temperature of the hot liquid?
A glass flask of volume 200 cm³ is completely filled with mercury at 20°C. How much mercury will overflow when the temperature is raised to 100°C? (Given: γ_mercury = 1.82 × 10⁻⁴ /°C, γ_glass = 2.7 × 10⁻⁵ /°C)
A pendulum clock keeps correct time at 20°C. The length of the pendulum is 1 m and α = 1.2 × 10⁻⁵ /°C. How many seconds will the clock lose or gain per day when the temperature rises to 40°C?
Sample Questions
100 g of ice at 0°C is mixed with 100 g of steam at 100°C. What is the final temperature and composition of the mixture? (Latent heat of ice = 80 cal/g, Latent heat of steam = 540 cal/g, specific heat of water = 1 cal/g°C)
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100°C; mixture contains only water
Step 1: Heat required to convert 100 g ice at 0°C to water at 100°C = 100×80 + 100×1×100 = 8000 + 10000 = 18000 cal. Step 2: Heat available from condensing 100 g steam at 100°C = 100 × 540 = 54000 cal. Step 3: Since heat available (54000 cal) > heat required (18000 cal), all ice converts to water at 100°C. Remaining heat = 54000 − 18000 = 36000 cal after condensing some steam. Let x g of steam condense: x × 540 = 18000, x = 33.3 g. So 33.3 g steam condenses and all ice melts to water at 100°C. Final state: 100°C, all water. Step 4: All steam (100 g) gives 54000 cal but only 18000 cal is needed
A copper calorimeter of mass 100 g contains 200 g of water at 20°C. A piece of metal of mass 150 g at 80°C is dropped into the calorimeter. The final temperature is 25°C. If specific heat of copper = 0.1 cal/g°C and water = 1 cal/g°C, find the specific heat of the metal.
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0.1273 cal/g°C
Step 1: Heat gained by water = 200 × 1 × (25 − 20) = 1000 cal. Step 2: Heat gained by calorimeter = 100 × 0.1 × (25 − 20) = 50 cal. Step 3: Total heat gained = 1000 + 50 = 1050 cal. Step 4: By principle of calorimetry, heat lost by metal = heat gained. 150 × c × (80 − 25) = 1050. 150 × c × 55 = 1050. c = 1050 / 8250 = 0.1273 cal/g°C. Option B doubles the answer by using ΔT = 27.5 for metal. Option C halves the answer. Always account for calorimeter's heat gain.
In an experiment, 10 g of a solid at 100°C is dropped into a calorimeter containing 50 g of water at 20°C. The water equivalent of the calorimeter is 5 g. The final temperature is 30°C. What is the specific heat of the solid?
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0.786 cal/g°C
Step 1: Heat gained = (mass of water + water equivalent of calorimeter) × specific heat of water × rise in temp = (50 + 5) × 1 × (30 − 20) = 55 × 10 = 550 cal. Step 2: Heat lost by solid = m × c × ΔT = 10 × c × (100 − 30) = 700c. Step 3: By principle of calorimetry: 700c = 550. c = 550/700 = 0.786 cal/g°C. Option B forgets the water equivalent. Option C uses (100−20) instead of (100−30) for solid's temperature drop. Option D uses ΔT = 140 instead of 70.
The coefficient of volume expansion of an ideal gas at constant pressure is γ_P and at constant volume is γ_V. Which of the following statements is correct regarding an ideal gas at room temperature (300 K)?
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γ_P = γ_V = 1/T = 3.33 × 10⁻³ K⁻¹
Step 1: From ideal gas equation PV = μRT. At constant pressure: ΔV/V = ΔT/T, so γ_P = ΔV/(VΔT) = 1/T. Step 2: At constant volume: ΔP/P = ΔT/T, so γ_V = ΔP/(PΔT) = 1/T. Step 3: Both are equal to 1/T. At T = 300 K: γ_P = γ_V = 1/300 = 3.33 × 10⁻³ K⁻¹. Step 4: This shows that for ideal gases, both expansion coefficients are equal and DEPEND on temperature (they decrease as T increases). Option B is wrong — there is no reason for γ_P > γ_V. Option D is wrong because γ depends on T.
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