A73.0 g piece of metal with specific heat 0.622 jg∘c and at 105∘c is placed in 300. g of water at 27.0∘c. what will be the final temperature of the water?

29.7

Explanation:

Recall that the First Law of Thermodynamics demands that the total internal energy of an isolated system must remain constant. Any amount of energy lost by the metal must be gained by the water.  Therefore, heat given off by the metal = −heat taken in by water, or:

The equation used to calculate the quantity of heat energy exchanged in this process is:

Heat stops flowing when the two samples are at the same temperature, so same final temperature of the water will be the final temperature of the metal as well.  

Substitute in the known values for the equation above and rearrange to solve for T.

−(0.622Jg∘C)(73.0 g)(T−105.0∘C)=(4.184Jg∘C)(300. g)(T−27.0∘C)

Simplify by multiplying specific heat and mass.

−(45.406J∘C)(T−105.0∘C)=(1,255.2J∘C)(T−27.0∘C)

Then distribute the heat capacities (calculated in the previous step) to the temperature differences.

−(45.406TfJ∘C)+4,767.63J=(1,255.2TfJ∘C)−33,890J

Combine like terms.

−1,300TfJ∘C=−38,657J

T=29.72∘C

The answer should have three significant figures so round to 29.7∘C.  

=29.72∘C