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

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 capacity.

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.

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.

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.

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.

Step 1: Heat lost by hot water = heat gained by cold water Step 2: m1 c (T1 - Tf) = m2 c (T2 - Tf) Step 3: 50 x (80 - Tf) = 100 x (Tf - 20) Step 4: 50 x 80 - 50Tf = 100Tf - 2000 Step 5: 4000 + 2000 = 100Tf + 50Tf = 150Tf Step 6: Tf = 6000/150 = 40C Answer: Final temperature is 40C. Hot water cools from 80C to 40C, cold water warms from 20C to 40C.

Step 1: Given m = 2 kg, L(vaporization) = 2260000 J/kg Step 2: Formula: Q = mL Step 3: Q = 2 x 2260000 = 4520000 J Step 4: Q = 4520 kJ = 4.52 MJ Answer: 4520000 joules needed. This large amount of heat breaks all intermolecular bonds to form steam. Temperature stays constant at 100C throughout boiling.