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.
ScanSolve AI Answer
Welcome back bilgate — missed you this week. Here are the solutions for the questions from the image: 7. (e) A mass of 5 kg of water is heated with a 500 W heater for 20 minutes. (i) Calculate the heat supplied by the heater supplied during this time. Step 1: Convert time from minutes to seconds. t = 20 minutes × 60 s/minute = 1200 s Step 2: Use the formula for heat supplied by a heater, Q = P × t. Q = 500 W × 1200 s Q = 600000 J The heat supplied by the heater is 600000 J. (ii) Given that the specific heat capacity of water is 4200 J kg^-1 K^-1, determine the change in temperature of the water during this time. Step 1: Use the formula for heat absorbed, Q = mc T. We have Q = 600000 J (from part i), m = 5 kg, and c = 4200 J kg^-1 K^-1. Step 2: Rearrange the formula to solve for T. T = (Q)/(mc) T = 600000 J5 kg × 4200 J kg^-1 K^-1 T = (600000)/(21000) K T ≈ 28.57 K The change in temperature of the water is approximately 28.57 K. (iii) Explain why water is often used as a cooling liquid in car radiators. Water is often used as a cooling liquid in car radiators because it has a high specific heat capacity. This means it can absorb a large amount of heat energy from the engine without a significant rise in its own temperature, effectively cooling the engine. 7. (f) A student used an un-calibrated thermometer in the laboratory to measure the temperature of a given liquid. He first put it in pure melting ice and the mercury thread was 10 cm long. He then removed it and put it in steam from boiling water, the mercury thread was 50 cm long. Then he puts it in the liquid and the mercury thread was 35 cm long. (i) Define temperature. Temperature is a measure of the average kinetic energy of the particles (atoms or molecules) within a substance, indicating its degree of hotness or coldness. (ii) State two properties that make mercury suitable for use in thermometers. 1. Mercury has a uniform and visible expansion over a wide range of temperatures. 2. It has a high boiling point and a low freezing point, allowing it to measure a broad range of temperatures in liquid form. 3. It is a good conductor of heat, allowing it to quickly respond to temperature changes. (Any two of these are suitable). (iii) Calculate the temperature of the liquid. Step 1: Identify the given lengths and corresponding temperatures. Length at 0^ C (melting ice), L_0 = 10 cm. Length at 100^ C (boiling water), L_100 = 50 cm. Length in the liquid, L_x = 35 cm. Step 2: Use the formula for calculating temperature with an un-calibrated thermometer. T_x = (L_x - L_0)/(L_100) - L_0 × 100^ C Step 3: Substitute the values and calculate T_x. T_x = 35 cm - 10 cm50 cm - 10 cm × 100^ C T_x = 25 cm40 cm × 100^ C T_x = (5)/(8) × 100^ C T_x = 0.625 × 100^ C T_x = 62.5^ C The temperature of the liquid is 62.5^ C. Send me the next one 📸