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A 10.0-gram sample of H2O(l) at 23.0°C absorbs 209 joules of heat. What is the final temperature of the H2O(l) sample?

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1 Answer

Hi Sally,
This question deals with "heat capacity". For some reading, see for example
From the references, we find that the "specific heat capacity" (given the symbol cp) of water is about 4.18 J/(gram*K). This means that it takes about 4.18 joules of heat to raise the temperature of one gram of water by one degree Kelvin (which is the same as by one degree Centigrade). If we let ΔT represent the temperature change of the water, we can write the equation as
   ΔT = (heat available in joules) / (cp * mass in grams)
I'll let you finish it for us -- insert the values of heat available, cp, and mass of water into the equation; calculate ΔT; add this temperature change to the initial temperature, and you have the final temperature.
Good luck !