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{"ops":[{"insert":"\nAt the triple point of water the pressure of a fixed mass of gas is 2680 Pa. The temperature is changed to T while the volume of the gas is kept constant. The pressure is then 4870 Pa. i. Find the value of T. ii. What is the advantage of making this determination at such a low pressure?\n\n\nA platinum resistance thermometer is calibrated in ice and steam. It is then used to find the temperature of an oil bath. The values of its resistance are: At the ice point: 25.6 \u03a9 At the steam point: 35.6 \u03a9 At \n\n\nA fixed mass of gas at constant pressure has a volume of 200 cm3 at the ice point, 273.2 cm3 at the steam point and 525.1 cm3 at the melting point of a solid. What is the melting point of the solid on the constant pressure gas scale of temperature?\n\n\nThe resistance of a resistance thermometer at the ice point is 40 \u03a9. Its resistance at temperatures T1 and T2 are 48 \u03a9 and 53.2 \u03a9 respectively. What is the ratio of T2 and T1?\u00a0\n\n\n\nA hot liquid at 80oC is added to 600 g of the same liquid originally at 10oC. When the mixture reaches 30oC, the total mass of liquid is?\n\nA mercury-in-glass thermometer is to be used to measure the temperature of some oil. The oil has a mass of 32.0 g and a specific heat capacity of 1.40 J g\u20131K\u20131. The actual temperature of the oil is 54.0 \u00b0C. The bulb of the thermometer has a mass of 12.0 g and an average specific heat capacity of 0.180 J g\u20131K\u20131. Before immersing the bulb in the oil, the thermometer reads 19.0 \u00b0C. The thermometer bulb is placed in the oil and the steady reading on the thermometer is taken. Determine i. The steady temperature recorded on the thermometer ii. The ratio of change in temperature of the oil to the initial temperature of the oil. iii. Suggest, with an explanation, a type of thermometer that would be likely to give a smaller value for the ratio calculated.\n\n\nGood day Sir, Thank you for helping others understand physics. I hope you would help me understand physics because I love it so much but due to some personal issues it has been very hard to understanding physics. Thank you very much for helping me out.\n"}]}

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