Question

Not sure how the final answer was obtained, for the Tfinal of this previously answered question. http://www.chegg.com/homework-help/questions-and-answers/mixture-containing-214g-ice-exactly-000-degrees-celsius-753g-water-553-degrees-celsius-pla-q5337662

Answer 1

A mixture containing 21.4g of ice (at exactly 0.00 degrees celsius) and 75.3g of water at (55.3 degrees celsius) is placed in an insulated container. Assuming no loss of heat to the surroundings, what is the final temperature of the mixture?

We use heat capacity of ice and liquid water.

C (heat capacity of ice ) = 2.03 J / g C

C (water ) = 4.184 J / deg C g

We know heat taken by ice would be given by water.

q (ice) = - q (water)

we know q = m C delta T

here , q is heat exchanged, C is specific heat , delta T is change in temperature,

Lets plug the corresponding values in above equation.

( 21.4 x 2.03 x (Tf – 0 ) = -( 75.3 x 4.184 x ( Tf-55.3 )

Here we have to find the value of Tf which will be the final temperature of the mixture.

21.4 x 2.03 x Tf = -( 75.3 x 4.184 x Tf-55.3 )

Tf = 48.6 C

So the final temperature of the mixture would be 64.1