Question 1: Define specific heat capacity and explain its significance in heat transfer.

Answer: Specific heat capacity is the amount of heat energy needed to raise the temperature of 1 kilogram of a substance by 1 degree Celsius (or Kelvin). It is significant because it determines how a material responds to heat energy, influencing heating and cooling processes.

Question 1: What units are used to express specific heat capacity?

Answer: Joules per kilogram per degree Celsius (J/kg°C) or Joules per kilogram per Kelvin (J/kg·K).

Question 2: Describe the mechanism by which heat energy increases the temperature of a substance.

Answer: Heat energy is transferred to the substance, increasing the kinetic energy of its particles, which results in a rise in temperature.

Question 1: A 2 kg block of aluminium absorbs 6000 Joules of heat energy, causing its temperature to rise by 10°C. Calculate the specific heat capacity of aluminium.

Answer: c = Q / (m × ΔT) = 6000 J / (2 kg × 10°C) = 300 J/kg°C.

Question 2: If 500 Joules of heat is supplied to 0.5 kg of copper and its temperature increases by 5°C, what is the specific heat capacity of copper?

Answer: c = Q / (m × ΔT) = 500 J / (0.5 kg × 5°C) = 200 J/kg°C.

Question 1: Describe a basic experimental setup to determine the specific heat capacity of a metal, including the variables you would control.

Answer: The experiment involves heating a metal sample, measuring its initial temperature, then transferring it to water and recording the final temperature. The variables include mass of the metal, initial temperature, heat energy supplied, and water temperature. Variables to control include consistent heating source, insulation to prevent heat loss, and accurate measurements of mass and temperature.

Question 2: List two safety precautions to consider when conducting experiments involving heating metals.

Answer: Wear heat-resistant gloves and safety goggles to protect against burns and splashes of hot metal or water.

Question 1: A 3 kg block of lead absorbs 18,000 Joules of heat, raising its temperature by 15°C. Calculate its specific heat capacity and compare it with that of aluminium. What does this tell you about lead’s thermal properties?

Answer: c = Q / (m × ΔT) = 18000 J / (3 kg × 15°C) = 400 J/kg°C. Lead’s specific heat capacity is higher than aluminium’s (300 J/kg°C), indicating it heats up more slowly and retains heat longer.

Question 1: Explain why materials with high specific heat capacities are used in building materials for insulation and temperature regulation.

Answer: Materials with high specific heat capacities absorb and store more heat without a large temperature increase, helping to regulate indoor temperatures and improve insulation efficiency.

Question 1: A substance has a specific heat capacity of 900 J/kg°C. If 2700 Joules of heat are added to 3 kg of the substance, by how much will its temperature increase? Show all your working.

Answer: Q = mcΔT Rearranged: ΔT = Q / (mc) = 2700 J / (3 kg × 900 J/kg°C) = 1°C. The temperature increases by 1 degree Celsius.