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On last Monday's lecture, Prof Lavelle did a problem solving calorimetry example, and I am confused about the way that the significant figures were calculated. The given temperatures were 25 and 31.9 degrees Celsius, and the delta T calculated was 6.9 degrees, but. it was my understanding that sig figs for subtraction were determined by the component with the highest degree of doubt; which means that the 25 number without a decimal would make the final answer 7 rather than 6.9, with only one sig fig. The final answer given was -2.9 kJ, with 2 sf.
Your rationale for sig figs is correct! But I believe that when Dr. Lavelle did this in class example, he wanted to focus more on the method for solving calorimetry heat transfer types of problems than being too picky about sig fig rules. On a test I think the sig figs of the given values would likely be more precisely and deliberately chosen.