# I know the answer, need help though. :3

## Homework Statement

An ice chest at a beach party contains 12 cans of soda at 3.35 °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 9.89-kg watermelon at 26.4 °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.

Side-note: I believe the the specific heat capacity of water is 4186 J/(kg C°).

q=cm∆T

## The Attempt at a Solution

I'm at a loss at how to attempt this equation. I know the answer is 20°C but I really don't understand how to use the equation to obtain it.

My attempt is that heat loss = heat gained but I haven't been able to work it out for myself.

Qlost=Qgained
MC∆T(watermelon)=MC∆T(soda)
9.89x4186x(26.4-t)=(0.35x12)x3800x(t-3.35)
1092947.856-41399.54t=15960t-53466
57359.54t=1146413.856
t=19.98645484

so the final temperature of the system will be approximately 20°C