An ice chest at a beach party contains 12 cans of soda at 2.45 °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 9.60-kg watermelon at 25.3 °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.
To determine the final temperature of the soda and watermelon, we can use the principle of conservation of energy. The heat gained by the watermelon will be equal to the heat lost by the soda.
We can start by finding the initial and final energies of the system.
The initial energy of the soda can be calculated using the equation: Q_initial = mass x specific heat capacity x change in temperature.
Q_initial_soda = 12 cans x 0.35 kg/can x 3800 J/(kg C°) x (T_initial_soda - 2.45 °C)
The initial energy of the watermelon is given by Q_initial_watermelon = mass x specific heat capacity x change in temperature.
Q_initial_watermelon = 9.60 kg x 4186 J/(kg C°) x (T_initial_watermelon - 25.3 °C)
The final energy of the system will be the sum of the initial energies, since heat is conserved.
Q_final = Q_initial_soda + Q_initial_watermelon
Now, we can set up and solve the equation:
Q_final = Q_initial_soda + Q_initial_watermelon
0 = 12 cans x 0.35 kg/can x 3800 J/(kg C°) x (T_final - 2.45 °C) + 9.60 kg x 4186 J/(kg C°) x (T_final - 25.3 °C)
Simplifying the equation:
0 = 12 cans x 0.35 kg/can x 3800 J/(kg C°) x T_final - 12 cans x 0.35 kg/can x 3800 J/(kg C°) x 2.45 °C + 9.60 kg x 4186 J/(kg C°) x T_final - 9.60 kg x 4186 J/(kg C°) x 25.3 °C
Now, we can combine the terms with T_final and those without T_final to solve for it:
12 cans x 0.35 kg/can x 3800 J/(kg C°) x T_final + 9.60 kg x 4186 J/(kg C°) x T_final = 12 cans x 0.35 kg/can x 3800 J/(kg C°) x 2.45 °C + 9.60 kg x 4186 J/(kg ° C°) x 25.3 °C
Now, we can solve this equation to find the value of T_final.