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Heat — Maharashtra (SSC) Class 10 Science Solutions (Free)

Free step-by-step Maharashtra (SSC) Class 10 Science solutions for "Heat" — important questions with detailed answers, download PDF for board exam preparation.

TL;DR: Free step-by-step Maharashtra (SSC) Class 10 Science solutions for "Heat" — important questions with detailed answers, download PDF for board exam pre…

By Syllab.in · Updated Jun 14, 2026

Q1: Define heat and temperature. Distinguish between them.

Heat: Form of energy transferred between bodies at different temperatures. Flows from hot to cold object. Unit: Joule (J) or calorie. It's energy in transit. Temperature: Measure of average kinetic energy of particles in a substance. Property of matter. Unit: Kelvin (K), Celsius (C), Fahrenheit (F). Difference: Heat is transferred energy, temperature is measure of molecular motion. Same amount of heat can change temperature differently for different substances based on their specific heat capaci…

Q2: Define specific heat capacity. How much heat is needed to raise 2 kg water from 20C to 30C?

Specific heat capacity: Heat required to raise temperature of 1 kg substance by 1 C. Unit: J/kg C. For water = 4200 J/kg C. Calculation: Q = mcDeltaT, where m = 2 kg, c = 4200 J/kg C, DeltaT = 10 C Q = 2 x 4200 x 10 = 84000 J = 84 kJ. Water has high specific heat, so it absorbs large amount of heat with small temperature rise. This is why water is good coolant.

Q3: Explain latent heat of fusion and latent heat of vaporization.

Latent heat of fusion: Heat required to convert 1 kg solid to liquid at constant temperature (melting point). For ice = 336000 J/kg. No temperature change during melting. Latent heat of vaporization: Heat required to convert 1 kg liquid to gas at constant temperature (boiling point). For water = 2260000 J/kg. No temperature change during boiling. During state change, all heat goes into breaking intermolecular bonds, not increasing kinetic energy, so temperature remains constant.

Q4: How much heat is needed to melt 5 kg ice at 0C? (Latent heat of fusion = 336000 J/kg)

Step 1: Given m = 5 kg, latent heat of fusion L = 336000 J/kg Step 2: Formula: Q = mL Step 3: Q = 5 x 336000 = 1680000 J Step 4: Q = 1680 kJ = 1.68 MJ Answer: 1680000 joules of heat is needed. This heat only melts ice; temperature remains 0C throughout melting.

Q5: Define calorimetry. Explain principle of conservation of energy in calorimetry.

Calorimetry: Method of measuring heat absorbed or released in a process using a calorimeter. Principle: Heat lost by hot object = heat gained by cold object (in isolated system with no heat loss to surroundings). Equation: Heat lost = Heat gained. If hot water at temp T1 mixes with cold water at temp T2, final temperature Tf is found by: m1 c (T1 - Tf) = m2 c (T2 - Tf). Assumes no heat loss to surroundings and no change in container.

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