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by dividing the qfreaction) by the mass of the soluble salt in grams, and is reported in units of joules/(gram of sample). The heat of reaction, in this case called the heat of solution, ??(reaction) is obtained AH(solution) q(reaction)/mass of salt The temperature difference is defined as follows ?T .. Tfinal-Tinitial Remember that there are 100 marks between the freezing point of water and the boiling point of water on a thermometer when marked in Â°C. There are also 100 marks between the freezing point of water and the boiling point of water on a thermometer marked in Kelvin. AT (between freezing and boiling point of water in Â°C) -100 Â°C AT (between freezing and boiling point of water in K) - 100 K Therefore, AT is the same numerical value whether in Â°C or K. 5. IfAT represents a DIFFERENCE in temperature, how does ?? (in Â°C) relate to AT (in K)? Write a simple equation that represents this relationship. Solution The simple equation can be delta T ( o C) = delta T (K) (where delta is the triangle symbol) delta T has the same numerical value whether it is measured in degree celcius or in Kelvin. The relationship to change Temperature in degree celcius to Kelvin is : T ( o C) = 273 + T (K) So T 1 ( o C) = 273 + T 1 (K) and T 2 ( o C) = 273 + T 2 (K) Now delta T = T 2 ( o C) - T 1 ( o C) ...(1) or delta T = [273 + T 2 (K)] - [273 + T 1 (K)] or delta T = 273 + T 2 (K) - 273 - T 1 (K) or delta T = T 2 (K) - T 1 (K) ...(2) From equation (1) and (2) T 2 ( o C) - T 1 ( o C) = T 2 (K) - T 1 (K) or delta T ( o C) = delta T (K) .

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