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.
I came out with Tf=25.18degreesC..But the answer is incorrect. Can someone please help?
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
To solve this problem, we can use the principle of conservation of energy, which states that energy is conserved in an isolated system.
First, let's calculate the initial energy of the system. The energy E can be calculated using the formula:
E = mass * specific heat capacity * temperature
For the 12 cans of soda:
E_soda = 12 * 0.35 kg * 3800 J/(kg °C) * 2.69 °C
E_soda = 11544 J/°C
For the watermelon:
E_watermelon = 6.05 kg * 3800 J/(kg °C) * 24.1 °C
E_watermelon = 55435.8 J/°C
Now, when the watermelon is added to the ice chest, heat transfer occurs until the final temperature T is reached. Since energy is conserved, we can equate the initial energy of the system to the final energy:
E_initial = E_final
E_soda + E_watermelon = (mass_soda + mass_watermelon) * specific heat capacity * T
Plugging in the values, we have:
11544 J/°C + 55435.8 J/°C = (12 * 0.35 kg + 6.05 kg) * 3800 J/(kg °C) * T
Simplifying the equation:
66979.8 J/°C = 9.2 kg * 3800 J/(kg °C) * T
Now, divide both sides of the equation by (9.2 kg * 3800 J/(kg °C)):
T = 66979.8 J/°C / (9.2 kg * 3800 J/(kg °C))
T ≈ 2.94 °C
Therefore, the final temperature of the soda and watermelon is approximately 2.94 °C.