What is the Final Temperature of Soda and Watermelon in the Ice Chest?

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AI Thread Summary
The discussion focuses on calculating the final temperature of soda and a watermelon in an ice chest using the principle of heat transfer. The initial temperatures and masses of both the soda and watermelon are provided, along with their specific heat capacities. The equation q=cm∆T is applied to equate the heat lost by the watermelon to the heat gained by the soda. After solving the equation, the final temperature is determined to be approximately 20°C. The calculations illustrate the concept of thermal equilibrium in a closed system.
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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°).

Homework Equations


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.
 
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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
 
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