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?

Question

Find the final temperature go the H20(l) sample.

Answer 1

Hi Sally,

 

This question deals with "heat capacity". For some reading, see for example

 

http://en.wikipedia.org/wiki/Heat_capacity

 

http://en.wikipedia.org/wiki/Water_(molecule)#Heat_capacity_and_heats_of_vaporization_and_fusion

 

From the references, we find that the "specific heat capacity" (given the symbol c_p) 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) / (c_p * mass in grams)

 

I'll let you finish it for us -- insert the values of heat available, c_p, 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 !