Question

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.

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