Questions: Thermal Properties of Matter

Q1. Why is the triple point of water a better fixed point than the melting point of ice?

The triple point is the unique temperature and pressure at which ice, liquid water and water vapour coexist in equilibrium. It has a single sharply defined value, 273.16 K, that does not depend on the environment or on how the apparatus is handled, because all three phases are present and the system cannot drift in temperature until one phase disappears. The melting point, by contrast, is the temperature at which ice melts at a given external pressure; changing the atmospheric pressure shifts it slightly, so two laboratories at different altitudes would get slightly different values. That is why modern thermometry uses the triple point as the primary fixed point.

Hinglish: Triple point par teeno phases ek saath hote hain aur temperature ka value 273.16 K fix rehta hai. Melting point pressure ke saath thoda badal jaata hai, isliye fixed point ke liye triple point zyada reliable hai.

Q2. State the ideal gas equation and use it to explain why a gas thermometer measures absolute temperature.

For a dilute gas, PV = nRT, so at constant volume the pressure is proportional to the absolute temperature: P = (nR/V) T. A constant-volume gas thermometer keeps the gas volume fixed and measures its pressure at the unknown temperature and at the triple point. The ratio of pressures equals the ratio of absolute temperatures, so T = 273.16 (P/P_tr). Because the proportionality involves the absolute temperature, not the Celsius temperature, the gas thermometer directly measures the Kelvin temperature. Real gases deviate slightly, so readings are extrapolated to zero pressure.

Hinglish: Constant volume par P absolute temperature ke proportional hota hai. Isliye pressure ka ratio temperature ka ratio de deta hai: T = 273.16 P/P_tr. Real gas ki chhoti deviation ko zero pressure par extrapolate karke hataate hain.

Q3. A steel rail is 20 m long at 20 degrees C. Find its length at 50 degrees C, with alpha = 1.2 x 10^-5 K^-1. How much gap must be left between rails?

The temperature rise is 30 K, so the expansion is L alpha delta T = 20 x 1.2 x 10^-5 x 30 = 7.2 x 10^-3 m, that is 7.2 mm. The new length is therefore 20.0072 m. If two rails butt together at 20 degrees C with no gap, the expansion on a hot day would make them press against each other and buckle the track, so a gap of at least about 7 mm is left. The gap is set at the lowest expected temperature so that the rails just touch at the highest expected temperature.

Hinglish: Delta L = L alpha delta T se 7.2 mm expansion aata hai. Isliye rails ke beech kam se kam 7 mm ka gap chhodna padta hai, warna garmi me track buckle ho sakta hai.

Q4. Distinguish between heat capacity and specific heat capacity, and give the unit of each.

Heat capacity is the heat needed to raise the temperature of a whole body by one kelvin, and it depends on the mass and the material of that body; its unit is J K^-1. Specific heat capacity is the heat needed to raise the temperature of unit mass of the material by one kelvin, so it is a property of the material only; its unit is J kg^-1 K^-1. The relation is Q = C delta T for the body and Q = mc delta T for a sample of mass m, with C = mc. In calorimetry problems the calorimeter is often described by its heat capacity W, which plays the same role as C in the equation.

Hinglish: Heat capacity poore body ke liye hoti hai aur mass par depend karti hai, unit J/K. Specific heat ek material ki property hai, unit J/kg/K. Relation simple hai: C = mc.

Q5. What is latent heat? Why does the temperature not change during melting or boiling?

Latent heat is the heat needed to change the state of unit mass of a substance at its transition temperature, with no change of temperature. During melting, the heat breaks the regular arrangement of the solid; during boiling, it supplies the energy needed to separate the molecules completely against the attractive forces. The average kinetic energy of the molecules therefore does not increase while the transition continues, so the thermometer reading stays constant. Once the whole sample has changed state, further heat again raises the temperature. For water, L_fusion is about 3.34 x 10^5 J kg^-1 and L_vaporisation about 22.6 x 10^5 J kg^-1.

Hinglish: Latent heat state badalne ke liye lagti hai. Melting me bond tootte hain aur boiling me molecules alag hote hain, isliye average kinetic energy nahi badhti aur temperature constant rehta hai. Poora sample badal jaane ke baad hi temperature phir badhta hai.

Q6. Explain conduction, convection and radiation with one example of each.

Conduction transfers heat through a material without moving the material, as when the handle of a metal spoon left in hot tea warms up. Convection transfers heat by the motion of a fluid, as when hot air rises above a room heater and cool air sinks. Radiation transfers heat by electromagnetic waves and needs no medium, as with the heat reaching the Earth from the Sun across empty space. A vacuum flask blocks all three: the double wall with a vacuum stops conduction and convection, and the silvered surfaces reduce radiation.

Hinglish: Conduction me material ke andar heat chalta hai, convection me fluid khud chalta hai, radiation me waves se heat aati hai aur medium ki zaroorat nahi. Thermos flask teenon ko rokta hai.

Q7. A 2 kg block of aluminium at 80 degrees C is placed in 1 kg of water at 20 degrees C. Find the final temperature, taking c_Al = 900 J kg^-1 K^-1 and c_w = 4200 J kg^-1 K^-1.

Heat lost by the aluminium equals heat gained by the water. Let T be the final temperature: 2 x 900 x (80 - T) = 1 x 4200 x (T - 20). So 1800(80 - T) = 4200(T - 20), which gives 144000 - 1800T = 4200T - 84000, and 6000T = 228000, so T = 38 degrees C. The final temperature lies between the two starting temperatures and closer to the water temperature because water has the larger total heat capacity in this example.

Hinglish: Heat lost = heat gained lagao. 1800(80 - T) = 4200(T - 20) solve karo, final temperature 38 degree Celsius aata hai. Paani ki heat capacity zyada hai, isliye final temperature paani ke paas hoti hai.

Q8. Derive the expression for the heat conducted through a slab in steady state and define thermal resistance.

In steady state the temperature at every point of the slab is constant in time, so the heat entering one face equals the heat leaving the other. Experiment shows the rate is proportional to the area, to the temperature difference and inversely proportional to the thickness: Q/t = k A (T1 - T2)/L. The constant k is the thermal conductivity. Comparing with Ohm's law, the quantity L/(kA) is called the thermal resistance of the slab, and the temperature difference is analogous to a potential difference. Slabs placed in series have the same heat current and their thermal resistances add, which lets composite walls and pipes be calculated slab by slab.

Hinglish: Steady state me rate kA(T1 - T2)/L hota hai. L/(kA) ko thermal resistance kehte hain, aur series me resistances add hote hain, jaise electrical circuit me.

Q9. State Newton's law of cooling and explain why the cooling curve is exponential.

When a body sits in surroundings at temperature T_s and its own temperature is only a little above T_s, the heat it gives up per second grows in proportion to the gap (T - T_s). Writing that loss as mc dT for a short interval dt gives mc dT/dt = -K(T - T_s). Separating the variables gives dT/(T - T_s) = -(K/mc) dt, and integrating gives T - T_s = (T_0 - T_s) e^-kt, an exponential fall. As the excess temperature falls, so does the rate, which is why the curve flattens out and never crosses the surroundings' temperature.

Hinglish: Cooling rate excess temperature ke proportional hota hai: mc dT/dt = -K(T - T_s). Integrate karo to exponential decay milta hai, T - T_s = (T0 - T_s) e^-kt. Jab excess kam hota hai to rate bhi kam hota hai, isliye curve flatten hota jaata hai.

Q10. Why does a bimetallic strip bend on heating, and how would you decide which metal is on the outside of the curve?

A bimetallic strip is made by joining two different metals. When heated, each metal tries to expand by its own amount, and because they are joined, the strip must bend. The metal with the larger coefficient of expansion becomes the longer outer arc, so it lies on the convex side; the metal with the smaller coefficient lies on the concave side. For a brass-steel strip, brass has the larger alpha, so on heating brass is on the outside. Bimetallic strips are used in thermostats and in old thermometers, where the bending switches a contact on or off at a set temperature.

Hinglish: Do metals alag alag expand karte hain, isliye joined strip mud jaati hai. Jiska alpha bada hota hai wo longer outer arc par, yaani convex side par hota hai. Brass-steel strip me heating par brass bahar hota hai.

Quiz below: five questions on temperature scales, expansion, calorimetry, latent heat and heat transfer. Try them before opening the formula sheet.