# An ice chest at a beach party contains 12 cans of soda at 3.05 °C. Each can of soda has a mass of 0.35 kg and a specific heat capacity of 3800 J/(kg C°). Someone adds a 7.15-kg watermelon at 25.0 °C to the chest. The specific heat capacity of watermelon is nearly the same as that of water. Ignore the specific heat capacity of the chest and determine the final temperature T of the soda and watermelon in degrees Celsius.

heat lost + heat gained = 0

mcdeltaT + mcdeltaT = 0

substitute Tf - Ti for delta T (in both cases). Tf is final T and Ti is initial T. Solve for Tf. Post your work if you get stuck.

Jena: Please don't change names on us. We try to get to know what kind of errors each student is making in order to effectively instruct. Changing names messes that up.

Thanks.

The sum of the heat gains is zero (one will gain heat, the other lose (negative gain)).

0=heatwatermelon gained + heatsodagained

0=masswatermellon*c*(Tf-25) + 12*.35*csoda*(Tf-3.05)

solve for Tf.

(.35)(3800)(Tf-3.05)+(7.15)(4186)(Tf-25)=0

OK but how do I get the Tf for both in the same time

I am Jena's sister

Sorry if I made an error.bobpursley

You haven't multiplied 0.35 kg/can x 12 cans = ?? kg soda.

Just do algebra and the Tf will be the only unknown in the equation.

I am stuck I got a whole different #.

12*0.35*3800*(Tf-3.05) + 7.15*4186*(Tf-25.0)=0

15960Tf - 48678 + 29929.9Tf - 748247.5 = 0

Solve for Tf. I assume you can take it from here.