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# specific heat capacity

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• 1. Present by: ‘izzat najmi bin ibrahimNoor fatinah binti muhd rusli Punitha A/P Nagappan
• 2. DEFINITITION A materials heat capacity is a measure of how much energy must be exchanged between an object and its environment to produce a change in temperature. In other words, it is a measure of an objects "capacity" to hold "heat." Materials with high heat capacities, such as water, require large amounts of energy to produce a small temperature change. Heat Capacity Heat capacity is the amount of heat required to increase the temperature of an object by 1 oC (or 1 K).
• 3. TYPE OF HEAT CAPACITY Molar heat capacity Specific heat Heat capacity
• 4. FORMULA OF SPECIFIC HEAT CAPACITY
• 5. EXAMPLEHow much thermal energy is required to raise the temperature of a 2 kg aluminium block from 25 C to 30 C? [The specific heat capacity of aluminium is 900 Jkg-1 oC-1]Answer:Mass, m = 2kg Specific heat capacity, c = 900 Jkg-1 oC-1 Temperature change, θ = 30 - 25 = 5 oCThermal energy required, Q = mcθ = (2)(900)(5) = 9000J.
• 6. CONVERSION OF ELECTRICAL ENERGY INTOTHERMAL ENERGY
• 7. EXAMPLE A lead shot of mass 5g is placed at the bottom of a vertical cylinder thatis 1m long and closed at both ends. The cylinder is inverted so that the shotfalls 1 m. By how much will the temperature of the shot increase if thisprocess is repeated 100 times? [The specific heat capacity of lead is130Jkg-1K-1] Answer: m = 5gh = 1m × 100 = 100mg = 10 ms-2c = 130Jkg-1K-1θ=? In this case, the energy conversion is from potential energy to heatenergy. We assume that all potential energy is converted into heat energy.Therefore mgh = mcθgh = cθ
• 8. CONVERSION OF GRAVITATIONAL ENERGY INTOTHERMAL ENERGY
• 9. CONVERSION OF KINETIC ENERGY INTOTHERMAL ENERGY
• 10. MIXING 2 LIQUID
• 11. EXAMPLEExample 4What will be the final temperature if 500 cm3 of water at 0 �C is added to 200cm3 of water at 90 �C? [Density of water = 1gcm-3]Answer:The density of water is 1g/cm3, which means the mass of 1 cm3 of water is equal to 1g.Let the final temperature = θ m1 = 500g = 0.5kg c1 = c θ1 = θ - 0 = θ m2 = 200g = 0.2kg c2 = c θ2 = 90 - θm1c1θ1 = m2c2θ2 (0.5) c ( θ ) = (0.2) c ( 90 - θ ) 0.5 θ = 18 - 0.2 θ 0.5 θ + 0.2 θ = 18 0.7 θ = 18 θ = 25.71 oC
• 12. THE END