An ice chest at a beach party contains 12 cans of soda at 4.05 °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 6.21-kg watermelon at 24.0 °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

Physics

Barbara Burke

25 October, 12:39

An ice chest at a beach party contains 12 cans of soda at 4.05 °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 6.21-kg watermelon at 24.0 °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.

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Frank Bartlett · Answer 1

Below is the solution:

Heat soda=heat melon

m1*cp1 * (t-t1) = m2*cp2 * (t2-t); cp2=cpwater

12*0.35*3800 * (t-5) = 6.5*4200 * (27-t)

15960 (t-5) = 27300 (27-t)

15960t-136500=737100-27300t

43260t=873600

t=873600/43260

t=20.19 deg celcius