An ice chest at a beach party contains 12 cans of soda at 2.69 °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 6.05-kg watermelon at 24.1 °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.

Question

I came out with Tf=25.18degreesC..But the answer is incorrect. Can someone please help?

Answer 1

Of course it is wrong, you found the final temp warmer than averything.

the sum of the heats gained is zero.
Heatgainedsoda+heatgainedwatermelon=0
12*.35*3800(Tf-2.69)+6.05*Cw(Tf-24.1)=0

for cw, assume the watermelon is water.

solve for Tf