This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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Here are the solutions to questions 10, 11, 12, and 13. 10. A glass flask of volume 100cm^3 is filled to the brim with liquid whose cubical coefficient of expansion is 1×10^-3 per ^. The flask and its liquid content are originally at 20^. Determine the volume of liquid which will overflow upon heating the flask to 50^ if the coefficient of linear expansion of glass is 8×10^-6 per ^. Step 1: Identify the given values and calculate the change in temperature. Initial volume, V_0 = 100 cm^3. Cubical coefficient of expansion of liquid, _liquid = 1 × 10^-3 per ^. Linear coefficient of expansion of glass, _glass = 8 × 10^-6 per ^. Initial temperature, T_1 = 20^. Final temperature, T_2 = 50^. Change in temperature, T = T_2 - T_1 = 50^ - 20^ = 30^. Step 2: Calculate the cubical coefficient of expansion for glass. The cubical coefficient of expansion for a solid is approximately three times its linear coefficient of expansion: _glass = 3 _glass = 3 × (8 × 10^-6 per ^) = 24 × 10^-6 per ^ Step 3: Calculate the change in volume for the liquid and the flask. The change in volume is given by V = V_0 T. For the liquid: V_liquid = V_0 _liquid T = (100 cm^3) × (1 × 10^-3 per ^) × (30^) V_liquid = 3 cm^3 For the flask: V_flask = V_0 _glass T = (100 cm^3) × (24 × 10^-6 per ^) × (30^) V_flask = 0.072 cm^3 Step 4: Determine the volume of liquid that overflows. The volume of liquid that overflows is the difference between the expansion of the liquid and the expansion of the flask. V_overflow = V_liquid - V_flask V_overflow = 3 cm^3 - 0.072 cm^3 V_overflow = 2.928 cm^3 The volume of liquid that will overflow is 2.928 cm^3. 11. 500cm^3 of water is to be heated from room temperature (28^) to 100^ in order to prepare hot cup of coffee. (a) What is the minimum heat required? (b) How long will it take to heat the water with a 1000W heating coil which has a heating efficiency of 70\%? Given constants: Density of water, _water = 1 g/cm^3 = 1000 kg/m^3. Specific heat capacity of water, c_water = 4186 J/kg^. a) What is the minimum heat required? Step 1: Calculate the mass of the water. Volume of water, V = 500 cm^3. m = V = (1 g/cm^3) × (500 cm^3) = 500 g = 0.5 kg Step 2: Calculate the change in temperature. Initial temperature, T_1 = 28^. Final temperature, T_2 = 100^. T = T_2 - T_1 = 100^ - 28^ = 72^ Step 3: Calculate the minimum heat required using Q = mc T. Q = (0.5 kg) × (4186 J/kg^) × (72^) Q = 150696 J The minimum heat required is 150696 J. b) How long will it take to heat the water with a 1000W heating coil which has a heating efficiency of 70\%? Step 1: Calculate the useful power of the heating coil. Power of heating coil, P = 1000 W. Efficiency, = 70\% = 0.70. P_useful = P = 0.70 × 1000 W = 700 W Step 2: Calculate the time taken using Q = P_useful t. t = (Q)/(P_useful) = 150696 J700 W t ≈ 215.28 s To convert to minutes: t = (215.28)/(60) min ≈ 3.59 min It will take approximately 215.28 s (or 3.59 min). 12. A 40kg metal slab at temperature 600^ is taken from a furnace and plunged into 300kg of oil originally at 25^. The final temperature of the oil/slab is 40^. Determine the specific heat capacity of the metal if that of oil is 2100J/kg.^. Step 1: Identify the given values. Mass of metal, m_metal = 40 kg. Initial temperature of metal, T_metal,1 = 600^. Mass of oil, m_oil = 300 kg. Initial temperature of oil, T_oil,1 = 25^. Final equilibrium temperature, T_f = 40^. Specific heat capacity of oil, c_oil = 2100 J/kg^. Step 2: Apply the principle of calorimetry (Heat lost by metal = Heat gained by oil). m_metal c_metal (T_metal,1 - T_f) = m_oil c_oil (T_f - T_oil,1) Step 3: Substitute the values into the equation. (40 kg) × c_metal × (600^ - 40^) = (300 kg) × (2100 J/kg^) × (40^ - 25^) 40 × c_metal × 560 = 300 × 2100 × 15 22400 × c_metal = 9450000 Step 4: Solve for c_metal. c_metal = (9450000)/(22400) c_metal = 421.875 J/kg^ The specific heat capacity of the metal is 421.875 J/kg^. 13. A 100g ice-block at -20^ is dumped into a thermally insulated container of water at 0^. How much water is frozen if the specific heat of ice is 2302 J/Kg.^ and latent heat of fusion is 3.35×10^5 J/Kg? Step 1: Identify the given values. Mass of ice, m_ice = 100 g = 0.1 kg. Initial temperature of ice, T_ice,1 = -20^. Initial temperature of water, T_water,1 = 0^. Specific heat capacity of ice, c_ice = 2302 J/kg^. Latent heat of fusion of ice, L_f = 3.35 × 10^5 J/kg. The problem implies that the final equilibrium temperature is 0^, and some water freezes. This means the heat gained by the ice to warm up to 0^ is supplied by the freezing of some of the water. Step 2: Calculate the heat gained by the ice to warm from -20^ to 0^. Q_ice,warm = m_ice c_ice (0^ - T_ice,1) Q_ice,warm = (0.1 kg) × (2302 J/kg^) × (0^ - (-20^)) Q_ice,warm = (0.1 kg) × (2302 J/kg^) × (20^) Q_ice,warm = 4604 J Step 3: Determine the mass of water that freezes