10 questions · Form 4 Physics Bab 4: Heat
An immersion heater supplies 2000 J of energy to 0.1 kg of a liquid, increasing its temperature from 25°C to 45°C. What is the specific heat capacity of the liquid?
Prefer reading to quizzing? All 10 questions are listed below with the answer and explanation under each one.
1. An immersion heater supplies 2000 J of energy to 0.1 kg of a liquid, increasing its temperature from 25°C to 45°C. What is the specific heat capacity of the liquid?
Answer: A
Q = m c Δθ => 2000 = 0.1 × c × (45 - 25) => 2000 = 2 c => c = 1000 J kg-1 °C-1.
2. A balloon contains 2.0 L of air at 300 K. If heated to 450 K at constant atmospheric pressure, what is the new volume of the balloon?
Answer: B
V1 / T1 = V2 / T2 => 2.0300 = V2 / 450 => V2 = 2.0 × 450300 = 3.0 L.
3. According to Charles's Law, what happens to the volume of a fixed mass of gas when its temperature in Kelvin is doubled at constant pressure?
Answer: B
Charles's Law states V ∝ T. Doubling the absolute temperature T directly doubles the volume V.
4. How much heat is released when 0.2 kg of steam at 100°C condenses into water at 100°C? (Specific latent heat of vaporisation of water = 2.26 × 106 J kg-1
Answer: A
Q = m l_v = 0.2 × (2.26 × 106 = 4.52 × 105 J.
5. Which graph correctly represents Boyle's Law for an ideal gas at constant temperature?
Answer: B
Since P ∝ 1V, a graph of P versus 1V yields a straight line passing through the origin.
6. A uncalibrated liquid-in-glass thermometer has a mercury column length of 4 cm at 0°C and 24 cm at 100°C. What temperature is recorded when the column length is 14 cm?
Answer: C
θ = [L_θ - L_0L_100 - L_0] × 100 = [14 - 424 - 4] × 100 = (1020) × 100 = 50°C.
7. A trapped gas in a cylinder has a volume of 0.04 m³ at a pressure of 100 kPa. If the gas is compressed at constant temperature to a volume of 0.01 m³, what is the new pressure?
Answer: C
P1 V1 = P2 V2 => 100 × 0.04 = P2 × 0.01 => P2 = 40.01 = 400 kPa.
8. A fixed mass of gas in a rigid container has a pressure of 1.2 × 105 Pa at 27°C. What will its pressure be at 127°C?
Answer: A
Convert temperatures to Kelvin: T1 = 27 + 273 = 300 K; T2 = 127 + 273 = 400 K. P1 / T1 = P2 / T2 => 1.2 × 105300 = P2 / 400 => P2 = 1.6 × 105 Pa.
9. What variable is kept constant in Gay-Lussac's Law?
Answer: D
Gay-Lussac's Law investigates the relationship between pressure and temperature for a fixed mass of gas at constant volume.
10. What is absolute zero temperature on the Celsius scale?
Answer: B
Absolute zero is 0 K, which corresponds to -273°C (-273.15°C).