Heat Questions and Answers – Page 2

Specific heat capacity is a fundamental thermodynamic property. Option A is CORRECT — Specific heat (c) is defined as the amount of heat required to raise the temperature of 1 kg of a substance by 1 Kelvin (or 1°C). Its SI unit is J/kg/K (Joule per kilogram per Kelvin) also written as J·kg⁻¹·K⁻¹. This is derived from formula c = Q/(mΔT) where Q=heat in joules m=mass in kg ΔT=temperature change in K. Water has highest specific heat: 4186 J/kg/K = 1 cal/g/°C. Option B is wrong — raising 1g by 1°C describes calorie definition (CGS system) not SI. The SI definition uses 1kg and 1K and the unit is J/kg/K not Joule alone. Option C is wrong — rate of heat flow per unit area is heat flux with unit W/m² not specific heat. This is related to thermal radiation and conduction studies. Option D is wrong — total heat content (enthalpy) is different from specific heat. Calorie is a CGS unit not SI unit of specific heat.

Optical Pyrometer measures very high temperatures by analyzing radiation from hot bodies. Option A (Stefan's law) is wrong — Stefan's law (E=σT⁴) relates total energy to T⁴. It is used by Total Radiation Pyrometers not Optical Pyrometers. Stefan's law involves total radiation of all wavelengths. Option B (Newton's cooling law) is wrong — Newton's cooling law describes rate of heat loss proportional to temperature difference. It applies to convective cooling not to pyrometry. Option C (Wien's radiation law) is CORRECT — Optical Pyrometer works on Wien's radiation law (Wien's distribution law) which describes spectral distribution of radiation from a black body. The optical pyrometer uses the disappearing filament method: a reference filament is compared with the brightness of the target at a specific wavelength. Option D (Planck's law) is wrong — Planck's law is the complete description of black body radiation. Optical pyrometers use Wien's approximation (valid at shorter wavelengths and higher temperatures) which is simpler for practical measurement.

The master conversion formula is one of the most important formulas in temperature measurement. The formula must account for this offset of 32. Option B (C/100 = (F-32)/180 = R/80 = (K-273)/100) is also correct mathematically (this is the unsimplified form) but the simplified standard form used in all competitive exams is option C. Option C (C/5 = (F-32)/9 = R/4 = (K-273)/5) is CORRECT — this is the standard simplified master conversion formula. It is derived from: Celsius has 100 divisions Fahrenheit has 180 divisions Reamur has 80 divisions Kelvin has 100 divisions (273 to 373). Simplified ratios: 100:180:80:100 = 5:9:4:5. Thus: C/5 = (F-32)/9 = R/4 = (K-273)/5. Option D (C = F-32 = R = K-273) is wrong — this would mean 0°C = 32-32=0°F (wrong 0°C=32°F) and 0°C = 0-273 = -273K (wrong 0°C=273K). This formula has no valid basis.

Pressure cooker is a brilliant application of pressure-boiling point relationship. Option A is wrong — microwave radiation is used by microwave ovens not pressure cookers. Pressure cookers use conventional stove heating with steam pressure. At standard pressure (1 atm) water boils at 100°C. In a sealed pressure cooker pressure builds up (typically 2 atm). This raises the boiling point to about 120-125°C. At this higher temperature food cooks 2-3 times faster because chemical reactions (protein denaturation starch gelatinization) in food occur much faster at higher temperatures. Option C is wrong — reducing boiling point makes water boil at LOWER temperature which would slow cooking. Everest water boils at ~70°C) making cooking slower. Option D is wrong — steam is not a chemical catalyst. Steam transfers heat efficiently but plays no catalytic role.

Air coolers and air conditioners differ fundamentally in what they control. Option A is wrong — air cooler does NOT control humidity like an AC. In fact an air cooler INCREASES humidity because it evaporates water into the air. An AC DECREASES humidity by condensing moisture on evaporator coils. An air cooler controls air speed (fan speed) but does NOT control temperature directly the way an AC does. An air cooler cools air by evaporating water (evaporative cooling) which lowers temperature somewhat but cannot achieve precise temperature control. It increases air movement but true temperature regulation is absent. Air conditioner controls both temperature and air speed precisely. Option C is wrong — unlike an AC an air cooler cannot precisely control temperature. It cools by evaporation only and cannot lower temperature to a set precise value in humid conditions. Option D is wrong — air cooler definitely controls air speed (it has a fan). It does not control humidity — it actually increases humidity.

This tests understanding of thermal conductivity and temperature perception. Option A is wrong — iron and wood are both at the same room temperature. There is no difference in actual temperature. The difference in sensation comes from rate of heat transfer, not actual temperature. Option B is CORRECT — iron has much higher thermal conductivity (~80 W/m/K) than wood (~0.1-0.2 W/m/K). When you touch cold iron, it rapidly conducts heat AWAY from your warm hand making your hand feel colder quickly. Wood conducts heat away very slowly so hand retains warmth longer. Both at same temperature but iron removes heat from hand much faster. Option C is wrong — wood is a POOR conductor (0.1-0.2 W/m/K) while iron is a GOOD conductor (80 W/m/K). Option D is wrong — specific heat of iron (0.11 cal/g°C) is actually LOWER than many substances. The sensation of cold comes from thermal conductivity (rate of heat flow) not specific heat capacity.

Water shows anomalous expansion between 0°C and 10°C. Option A is wrong — continuous volume increase would be normal thermal expansion. Water does NOT follow this from 0 to 4°C. From 0°C to 4°C water CONTRACTS (volume decreases, density increases). At 4°C water reaches its MAXIMUM DENSITY. From 4°C to 10°C water EXPANDS normally (volume increases, density decreases). This anomalous behavior is due to the hydrogen bonding structure of water breaking down as temperature rises from 0 to 4°C, allowing molecules to pack more closely. Option C is wrong — continuous decrease would mean water always contracts on heating which is false above 4°C. Option D is wrong — volume cannot remain constant as temperature changes in real liquids.

The Total Radiation Pyrometer collects all radiation from a hot body to measure temperature. Option A is wrong — Wien's law relates peak wavelength to temperature and is used by Optical Pyrometers, not Total Radiation Pyrometers. Option B is wrong — Newton's cooling law describes rate of heat loss by convection and has no role in pyrometry. Option C is wrong — Snell's law is about refraction of light at boundaries and is unrelated to temperature measurement. Option D is CORRECT — Total Radiation Pyrometer is based on Stefan-Boltzmann law: E = σT⁴. It collects all wavelengths of radiation from the hot body and focuses them on a thermocouple or bolometer. By measuring total energy received, temperature is calculated using Stefan's law. This is slightly wider than Optical Pyrometer (800°C to 2700°C). Stefan constant σ = 5.67×10⁻⁸ W/m²K⁴.

Understanding how a refrigerator's components affect its cooling is important for competitive exams. Option A (Cooling becomes more efficient) is incorrect — if the freezer fails, losing the primary cooling mechanism means cooling will worsen, not improve. The freezer plays a central role in the refrigeration cycle. Option B (Cooling remains unaffected) is incorrect — the freezer compartment is integral to the refrigeration cycle. If it fails, the entire cooling process is compromised. The refrigerant circulates through the evaporator coils in and around the freezer section. Option C (Cooling becomes very poor) is CORRECT — the image states: रेफ्रिजरेटर के संबंध में सत्य कथन है — यदि फ्रीजर इकट्ठी होती है, तो रेफ्रिजरेटर में शीतलन बुरी तरह प्रभावित हो जाता है. The freezer is the coldest part of the refrigerator and is essential for the overall refrigeration cycle. The evaporator coils (where refrigerant absorbs heat) are concentrated in and around the freezer. If the freezer fails to form ice or malfunctions, the heat absorption process is severely disrupted, leading to very poor cooling throughout the entire refrigerator. Option D (Temperature increases rapidly) is partially related to C but the image specifically says cooling becomes very poor (बुरी तरह प्रभावित), which implies poor cooling and gradual temperature rise, not a rapid increase necessarily.

The choice of refrigerant for domestic refrigerators depends on efficiency, safety, and environmental factors. Option A (Ammonia) is incorrect — ammonia is the refrigerant used in LARGE cold storage units and industrial refrigeration, not in domestic (household) refrigerators. Ammonia is toxic, pungent, and requires robust containment, making it unsuitable for home use. Option B (Nitrogen) is incorrect — nitrogen is an inert gas used as a cryogen (liquid nitrogen at -196°C) for specialized applications. It is not used as a cycling refrigerant in domestic refrigerators. Option C (CFC — Chlorofluorocarbon) is CORRECT — the image states: बड़े शीतगृह संयंत्र प्रशीतक के रूप में अमोनिया का उपयोग करते हैं, जबकि घरेलू रेफ्रिजरेटर में CFC (क्लोरो-फ्लोरो कार्बन) का उपयोग करते हैं. CFC (Freon, R-12) was the standard domestic refrigerant. Modern domestic refrigerators now use HFC (R-134a, R-410A) as a safer alternative, but CFC is the answer given in traditional exam content. Option D (Helium) is incorrect — helium is used as a cryogenic coolant in MRI machines and superconducting magnets at near absolute zero temperatures. It has no practical application as a domestic refrigerator refrigerant.